WAEC SSCE - Further Mathematics - 2021

A body of mass 15kg is placed on a smooth plane which is inclined at 60° to the horizontal. If the box is at rest,

calculate the normal reaction to the plane. [ Take g = 10m/s\(^2\) ]

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To find the normal reaction to the plane, we can use the following approach: Step 1: Draw a diagram of the situation. In this case, we have a body of mass 15kg on a smooth plane inclined at 60° to the horizontal. The weight of the body acts vertically downwards and is equal to mg, where m is the mass of the body and g is the acceleration due to gravity. Step 2: Resolve the weight of the body into two components, one perpendicular to the plane and the other parallel to the plane. The component of the weight perpendicular to the plane is equal to mg cos 60°, which is equal to (15 kg) x (10 m/s\(^2\)) x cos 60° = 75 N. Step 3: The normal reaction to the plane is equal in magnitude but opposite in direction to the component of the weight perpendicular to the plane. Therefore, the normal reaction to the plane is 75 N in the upward direction. Step 4: Check the options given in the question and select the one that matches the value obtained in Step 3. In this case, the correct option is 75N. Therefore, the normal reaction to the plane is 75 N.

If ( 1- 2x)\(^4\) = 1 + px + qx\(^2\) - 32x\(^3\) + 16\(^4\), find the value of (q - p)

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The first term of an AP is 4 and the sum of the first three terms is 18. Find the product of the first three terms

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Let the common difference of the AP be denoted by d. Then, the first three terms of the AP are 4, 4 + d, and 4 + 2d, respectively. The sum of the first three terms of the AP is given as 18. Therefore, we have: 4 + (4 + d) + (4 + 2d) = 18 Simplifying the above equation, we get: 3d + 12 = 18 3d = 6 d = 2 Hence, the common difference of the AP is 2. Therefore, the first three terms of the AP are: 4, 6, 8 The product of the first three terms is: 4 x 6 x 8 = 192 Therefore, the answer is 192.

If f(x) = 4x\(^3\) + px\(^2\) + 7x - 23 is divided by (2x -5), the remainder is 7. find the value of p

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Given that M = \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\) and N = \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\), calculate (3M - 2N)

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To calculate 3M - 2N, we need to multiply each matrix by its scalar factor and then subtract the results. First, we have: 3M = 3 x \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\) = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) Next, we have: 2N = 2 x \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\) = \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\) Now we can subtract these two matrices to get: 3M - 2N = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) - \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\) = \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\) Therefore, the correct answer is (B) \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\).

Simplify ( \(\frac{1}{2 - √3}\) + \(\frac{2}{2 + √3}\) )\(^{-1}\)

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Given that f: x --> x\(^2\) - x + 1 is defined on the Set Q = { x : 0 ≤ x < 20, x is a multiple of 5}. find the set of range of F.

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Solve (\(\frac{1}{9}\))\(^{x + 2}\) = 243\(^{x - 2}\) 

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Find the equation of the normal to the curve y= 2x\(^2\) - 5x + 10 at P(1, 7).

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If \(\frac{15 - 2x}{(x+4)(x-3)}\) = \(\frac{R}{(x+4)}\) \(\frac{9}{7(x-3)}\), find the value of R

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Using binomial expansion of ( 1 + x)\(^6\) = 1 + 6x + 15x\(^2\) + 20x\(^3\) + 6x\(^5\) + x)\(^6\), find, correct to three decimal places, the value of (1.998))\(^6\)

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To use the binomial expansion of (1 + x)\(^6\), we substitute x = 1.998, which gives: (1 + 1.998)\(^6\) = 1 + 6(1.998) + 15(1.998)\(^2\) + 20(1.998)\(^3\) + 6(1.998)\(^5\) + (1.998)\(^6\) We are interested in finding the value of (1.998)\(^6\), which is the last term on the right-hand side of the equation. We can solve for it by subtracting the other terms from both sides of the equation: (1.998)\(^6\) = (1 + 1.998)\(^6\) - 1 - 6(1.998) - 15(1.998)\(^2\) - 20(1.998)\(^3\) - 6(1.998)\(^5\) Using a calculator, we can evaluate the right-hand side of the equation to get: (1.998)\(^6\) ≈ 63.167 Therefore, the correct answer is option B, 63.167, rounded to three decimal places. Explanation: The binomial expansion of (1 + x)\(^6\) is a formula that allows us to expand the expression into a sum of terms involving powers of x. By substituting x = 1.998, we can use the formula to find the value of (1.998)\(^6\). We then use a calculator to evaluate the expression to obtain the final answer.

The gradient ofy= 3x\(^2\) + 11x + 7 at P(x.y) is -1. Find the coordinates of P. 

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If √5 cosx + √15sinx = 0, for 0° < x < 360°, find the values of x.

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A stone is thrown vertically upward and distance, S metres after t seconds is given by S = 12t + \(\frac{5}{2t^2}\) - t\(^3\).

Calculate the maximum height reached.

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Find the inverse of  \(\begin{pmatrix} 4 & 2 \\ -3 & -2 \end{pmatrix}\)

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If 2i +pj and 4i -2j are perpendicular, find the value of p.

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To find the value of p, we need to use the concept of perpendicular vectors. Two vectors are perpendicular if and only if the dot product of the two vectors is equal to zero. The dot product of two vectors can be calculated as the product of the magnitudes of the two vectors and the cosine of the angle between them. If the angle between the two vectors is 90 degrees, the cosine of the angle is zero and the dot product is also zero. Therefore, to find the value of p, we need to calculate the dot product of the two vectors, 2i + pj and 4i - 2j, and set it equal to zero. The dot product of two vectors (a, b) and (c, d) is given by: (a, b) * (c, d) = ac + bd So, the dot product of 2i + pj and 4i - 2j is: (2i + pj) * (4i - 2j) = (2 * 4) + (p * -2) = 8 - 2p = 0 Therefore, 2p = 8 and p = 4. So, the value of p is 4.

Find the value of the derivative of y = 3x\(^2\) (2x +1) with respect to x at the point x = 2. 

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A fair die is tossed 60 times and the results are recorded in the table

Find the probability of obtaining a prime number.

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A prime number is a number greater than 1 that is only divisible by 1 and itself. The prime numbers on a die are 2, 3, and 5. We know that the die is fair, which means that each number has an equal probability of being rolled. There are a total of 60 rolls, so we can use the frequency table to count the number of times each prime number appears on the die. The frequency of the number 2 is 10, the frequency of the number 3 is 14, and the frequency of the number 5 is 8. Therefore, the total number of times a prime number was rolled is: 10 + 14 + 8 = 32 The probability of rolling a prime number on a fair die is the total number of times a prime number was rolled divided by the total number of rolls: 32/60 = 8/15 Therefore, the answer is option D, 8/15.

For what value of k is 4x\(^2\) - 12x + k, a perfect square?

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A bag contains 8 red, 4 blue and 2 green identical balls. Two balls are drawn randomly from the bag without replacement. Find the probability that the balls drawn are red and blue.

A. 12/91 B. C. D.

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If α and β are the roots of 3x\(^2\) - 7x + 6 = 0, find \(\frac{1}{α}\) + \(\frac{1}{β}\)

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Given that F\(^1\)(x) = x\(^3\) √x, find f(x)

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In how many ways can 8 persons be seated on a bench if only three seats are available?

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The first person has 8 choices for a seat, the second person has 7 choices remaining, and the third person has 6 choices. However, the order in which they choose their seats does not matter, so we need to divide by the number of ways that the three people can be arranged. This is given by 3 factorial (3!), which is 3 x 2 x 1 = 6. Therefore, the total number of ways that 8 persons can be seated on a bench if only three seats are available is: 8 x 7 x 6 / 3! = 336 So the correct answer is (C) 336.

In △PQR, \(\overline{PQ}\) = 5i - 2j and \(\overline{QR}\) = 4i + 3j. Find \(\overline{RP}\).

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g(x) = 2x + 3 and f(x) = 3x\(^2\) - 2x + 4

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A stone is thrown vertically upward and distance, S metres after t seconds is given by S = 12t + \(\frac{5}{2t^2}\) - t\(^3\).

Calculate the distance travelled in the third second.

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Simplify \(\frac{1}{3}\) log8 + \(\frac{1}{3}\) log 64 - 2 log6

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The table shows the distribution of marks obtained by some students in a test

Find the modal class mark.

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To find the modal class mark, we need to first identify the class with the highest frequency, which in this case is the class with marks 20-29 (16 students). The modal class mark is the midpoint of this class, which is calculated by adding the lower and upper limits of the class and dividing by 2: Modal class mark = (20 + 29) / 2 = 24.5 Therefore, the modal class mark is 24.5, which is option (C).

Given that P = { x: 0 ≤ x ≤ 36, x is a factor of 36 divisible by 3} and Q = { x: 0 ≤ x ≤ 36, x is an even number and a perfect square}, find P n Q.

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A committee consists of 6 boys and 4 girls. In how many ways can a sub-committee consisting of 3 boys and 2 girls be formed if one particular boy and one particular girl must be on the sub-committee?

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For what range of values of x is x\(^2\) - 2x - 3 ≤ 0

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To solve the inequality x\(^2\) - 2x - 3 ≤ 0, we can use factoring or the quadratic formula. Factoring gives us (x - 3)(x + 1) ≤ 0, which means that the expression is less than or equal to zero when x is between or equal to -1 and 3, since the factors change sign at these values. Therefore, the correct answer is {x: -1 ≤ x ≤ 3}. Alternatively, we can use the quadratic formula to find the roots of the equation x\(^2\) - 2x - 3 = 0, which are x = -1 and x = 3. Since the quadratic function is a parabola that opens upward, it is negative in the interval between these two roots. Therefore, the expression x\(^2\) - 2x - 3 is less than or equal to zero when x is between or equal to -1 and 3.

Three forces, F\(_1\) (8N, 030°), F\(_\2) (10N, 150° ) and F\(_\3) ( KN, 240° )are in equilibrium. Find the value of N

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If sin x = \(\frac{12}{13}\) and sin y = \(\frac{4}{5}\), where x and y are acute angles, find  cos (x + y)

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If 2y\(^2\) + 7 = 3y - xy, find \(\frac{dy}{dx}\)

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A binary operation * is defined on the set of real numbers, R, by

P * q = \(\frac{q^2 - p^2}{2pq}\). Find 3 * 2

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The binary operation * is defined as: P * q = (q^2 - p^2) / (2pq). To find 3 * 2, we need to substitute the values of p and q in the equation: 3 * 2 = (2^2 - 3^2) / (2 * 3 * 2) = (-5) / (12) = -5/12 So, the answer is -5/12.

The table shows the distribution of marks obtained by some students in a test

What is the upper class boundary of the upper quartile class?

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Find the radius of the circle 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0. 

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To find the radius of a circle from its equation in general form, we need to rewrite the equation in the standard form of a circle, which is: (x - h)\(^2\) + (y - k)\(^2\) = r\(^2\) Where (h, k) is the center of the circle, and r is its radius. To do this, we complete the square for both x and y terms in the given equation. We start by rearranging the terms as follows: 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0 2x\(^2\) - 4x + 2y\(^2\) - 6y = 2 Now, we need to add and subtract appropriate constants to complete the square for x and y terms separately. For the x terms, we take half of the coefficient of x (-4/2 = -2) and square it to get 4. So, we add and subtract 4 to the equation: 2x\(^2\) - 4x + 4 - 4 + 2y\(^2\) - 6y = 2 We can now group the first three terms and factor it as a perfect square: 2(x - 1)\(^2\) + 2y\(^2\) - 6y - 2 = 0 For the y terms, we take half of the coefficient of y (-6/2 = -3) and square it to get 9. So, we add and subtract 9 to the equation: 2(x - 1)\(^2\) + 2(y - 3)\(^2\) - 2 - 9 = 0 2(x - 1)\(^2\) + 2(y - 3)\(^2\) = 11 Now, we have the equation in the standard form of a circle, where the center is at (1, 3) and the radius is the square root of 11/2. We can simplify this expression to get: sqrt(11/2) = sqrt(11)/sqrt(2) = sqrt(2)*sqrt(11)/2 Therefore, the answer is option (C), 17/√2.

Consider the following statements:

X: Benita is polite

y: Benita is neat

z: Benita is intelligent

Which of the following symbolizes the statement: "Benita is neat if and only if she is neither polite nor intelligent"?

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Evaluate: \(^9{?}_1\) \(\frac{x(2x-3)}{?x}\) dx

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9∫1 ${}^{}{<br>}_{}$ x(2x−3)√x $\frac{<br>}{}$ dx = 2x2−3x)x1/2 $\frac{{\mathrm{<br>}}^{}<br>}{{\mathrm{<span\; style="font-weight:\; var(--bs-body-font-weight);"><br></span>}}^{}}$

= x −1/2 ${}^{ \; \; \; }$ (2x2 ${}^2$ - 3x)

= 2x3/2 ${}^{<br>\; <span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}$ - 3x1/2 ${}^{<br>\; <span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}$

= 2x3/23/2+1 $\frac{{\mathrm{<br>}}^{\mathrm{<br>}}}{}$ - 3x1/21/2+1 $\frac{{\mathrm{<br>}}^{}}{\mathrm{<br>}}$

= 2x3/25/2 $\frac{{\mathrm{<br>}}^{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}}{}$ - 3x1/23/2 $\frac{{\mathrm{<br>}}^{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}}{}$

= 9∫1 ${}^{}{<br>}_{}$ [4x3/25 $\frac{{\mathrm{<br>}}^{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}}{}$ - 6x1/23 $\frac{{\mathrm{<br>}}^{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}}{}$] 

[ 4(9)3/25 $\frac{<br>}{}$- 6(9)1/23 $\frac{{<br>}^{}}{}$] - [ 4(9)1/25 $\frac{<br>}{}$ - 6(1)1/23 $\frac{<br>}{}$]

= [ 4(27)5 $\frac{<br>}{}$- 6(3)3 $\frac{<br>}{}$- [ 45 - 2 ]

=  [ 1085- 2 ] -  [ 65 ]

= 785 + 65= 845 $\frac{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br><br></span>}}{}$

164/5

A bag contains 24 mangoes out of which six are bad. If 6 mangoes are selected randomly from the bag with replacement, find the probability that not more than 3 are bad.

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The probability of selecting a bad mango from the bag is 6/24 = 1/4. This means that the probability of selecting a good mango is 3/4.

To find the probability that not more than 3 mangoes are bad, we need to consider four cases: selecting 0 bad mangoes, selecting 1 bad mango, selecting 2 bad mangoes, or selecting 3 bad mangoes.

Case 1: Selecting 0 bad mangoes

The probability of selecting a good mango on the first draw is 3/4. This probability remains the same for each of the 6 mangoes that are selected. Therefore, the probability of selecting 0 bad mangoes is (3/4)^6 = 0.1771.

Case 2: Selecting 1 bad mango

There are 6 ways to select 1 bad mango out of 6 mangoes, and 5 ways to select 5 good mangoes out of the remaining 18 mangoes. Therefore, the probability of selecting 1 bad mango is (6/24) * (18/24)^5 * 6 = 0.3934.

Case 3: Selecting 2 bad mangoes

There are 15 ways to select 2 bad mangoes out of 6 mangoes, and 4 ways to select 4 good mangoes out of the remaining 18 mangoes. Therefore, the probability of selecting 2 bad mangoes is (15/24)^2 * (9/24)^4 * 15 = 0.3147.

Case 4: Selecting 3 bad mangoes

There are 20 ways to select 3 bad mangoes out of 6 mangoes, and 3 ways to select 3 good mangoes out of the remaining 18 mangoes. Therefore, the probability of selecting 3 bad mangoes is (6/24)^3 * (18/24)^3 * 20 = 0.0983.

The total probability of not selecting more than 3 bad mangoes is the sum of the probabilities of these four cases: 0.1771 + 0.3934 + 0.3147 + 0.0983 = 0.9835.

Therefore, the probability of not more than 3 bad mangoes being selected is 0.9835 or approximately 98.35%.

\(^{5y}{C}_2\) = 190, find the value of y

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We can use the formula for combinations to solve for y:

^5yC₂ = ^(5y)! / _(2!(5y-2)!) = 190

Expanding the factorials:

^(5y)(5y-1) / 2 = 190

Solving for y:

(5y)(5y-1) = 380

25y^2 - 5y = 380

25y^2 - 5y - 380 = 0

Using the quadratic formula:

y = (-b ± √(b^2 - 4ac))/(2a)

y = (-(−5) ± √((-5)^2 - 4(25)(-380)))/(2(25))

y = (5 ± √(25 + 90000))/(50)

y = (5 ± √(90025))/(50)

Taking the positive root:

y = (5 + 300)/(50)

y = 305/50

y = 6.1

So the value of y is approximately 6.1.

(a) Find the equation of the normal to the curve y = (x\(^2\) - x + 1)(x - 2) at the point where the curve cuts the X - axis.

(b) The coordinates of the pints P, Q and R are (-1, 2), (5, 1) and (3, -4) respectively. Find the equation of the line joining Q and the midpoint of \(\overline{PR}\).

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(a) To find the equation of the normal to the curve y = (x² - x + 1)(x - 2) at the point where the curve cuts the X-axis, we first need to find the x-coordinate of the point where the curve intersects the X-axis. This is where y = 0, so we can solve the equation (x² - x + 1)(x - 2) = 0 to find the roots of the equation. The roots are x = 1 ± i√3 and x = 2, but since we want the point where the curve intersects the X-axis, we take x = 2.

Next, we need to find the gradient of the curve at this point. We can do this by differentiating the equation y = (x² - x + 1)(x - 2) with respect to x, giving:

dy/dx = 3x² - 8x + 1

Substituting x = 2, we get:

dy/dx = 3(2)² - 8(2) + 1 = -7

Therefore, the gradient of the curve at the point where it intersects the X-axis is -7.

Since the normal to the curve is perpendicular to the tangent at the point of intersection, we know that the gradient of the normal is the negative reciprocal of the gradient of the tangent. So the gradient of the normal is 1/7.

Finally, we can find the equation of the normal using the point-slope form of the equation of a straight line:

y - 0 = (1/7)(x - 2)

Simplifying this equation gives:

y = (1/7)x - 2/7

So the equation of the normal to the curve y = (x² - x + 1)(x - 2) at the point where the curve cuts the X-axis is y = (1/7)x - 2/7.

(b) To find the equation of the line joining Q and the midpoint of PR, we first need to find the coordinates of the midpoint of PR. The midpoint of a line segment with endpoints (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2). So the midpoint of PR is ((-1 + 3)/2, (2 - 4)/2), which simplifies to (1, -1).

Next, we need to find the gradient of the line joining Q and the midpoint of PR. We can use the gradient formula, which gives:

m = (y₂ - y₁)/(x₂ - x₁)

Substituting the coordinates of Q and the midpoint of PR gives:

m = (1 - (-1))/(5 - 1) = 1/2

(a) A girl threw a stone horizontally with a velocity of 30m/s from the top of a cliff 50m high. How far from the foot of the cliff does the stone strike the ground? [Take g= 10m/s\(^2\) 

(b) A body A, of mass 2kg is held in equilibrium by means of two strings AP and AR. AP is inclined at 56° to the upward vertical and AR is horizontal.

Find the tensions T\(_1\), and T\(_2\), in the strings [Take g= 10ms\(^2\)]

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(a)

Time taken to reach the ground

Using S = ut + 1/2gt2 ${}^2$

 (u= 0 for a body falling from rest)

50 = 0 x t + 1/2 x 10t2 ${}^2$

5t2 ${}^2$ = 50;

t2 ${}^2$ =50/5 = 10

 t = √10 =3.16

Distance when it strikes the ground This journey is independent of gravity

d =vt =30 x 3.16 = 94.8m

(b)

Using Lami's rule,

20sin34 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ = T1sin90 $\frac{{\mathrm{<br>\; <br>}}_{}}{\mathrm{<br>}}$ = T2sin56 $\frac{{\mathrm{<br>\; <br>}}_{}}{}$

= T1 ${}_1$ 20.sin90sin34 $\frac{\mathrm{<br>}\mathrm{<br>}}{\mathrm{<br>}}$ = T2 ${}_2$ 20.sin56sin34 $\frac{\mathrm{<br>}\mathrm{<br>}}{\mathrm{<br>}}$ 

T1 ${}_1$ = 35.79N and T2 ${}_2$ = 29.65N

The polynomial f(x) =2x\(^3\) + px+ qx - 5 has  (x-1) as a factor and a remainder of 27 when divided by (x + 2), where p and q are constants. Find the values of p and q.

2x^3 + px + qx - 5 
          ÷ x + 2
2          1    p/2    (q - 5)/2
-2x^2 + (p - 4)x + (q - 5)

-2x^2 + (p - 4)x + (q - 5) = 27

p - 4 = 0
p = 4
q - 5 = 27
q = 32

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To find the values of p and q, we can use the Remainder Theorem and synthetic division. The Remainder Theorem states that if a polynomial f(x) is divided by x - a, then the remainder is equal to f(a).

So, in this case, when we divide f(x) by (x - 1), the remainder is 27, meaning that f(1) = 27. We can use this information to find one of the constants, p.

Next, we can use synthetic division to divide f(x) by (x + 2) and find the remainder. The process of synthetic division involves dividing each term of the polynomial by the leading coefficient of the divisor, and then using the coefficients to find the remainder.

In this case, the polynomial f(x) is divided by (x + 2), giving us:

2x^3 + px + qx - 5 
          ÷ x + 2
2          1    p/2    (q - 5)/2
-2x^2 + (p - 4)x + (q - 5)

Since the remainder is 27, we have:

-2x^2 + (p - 4)x + (q - 5) = 27

Solving for p and q, we have:

p - 4 = 0
p = 4
q - 5 = 27
q = 32

So, the values of p and q are p = 4 and q = 32.

(a) A jogger is training for 15km charity race. He starts with a run of 500 metres, then he increases the distance he runs daily by 250 metres.

(i) How many days will it take the jogger to reach a distance of 15km in training?

(ii) Calculate the total distance he would have run in the training.

(b) The second term of a Geometric Progression (GP) is -3. If its sum to infinity is 25/2, find its common ratios.

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(a)

(i) To reach a distance of 15km, the jogger needs to cover 15000 meters.

Let's call the number of days it takes to reach this distance "n."

On the first day, he runs 500 meters. On the second day, he runs 500 + 250 = 750 meters. On the third day, he runs 750 + 250 = 1000 meters.

In general, on the nth day, he runs 500 + 250(n-1) meters.

So we can set up an equation:

500 + 750 + 1000 + ... + (500 + 250(n-1)) = 15000

Simplifying, we get:

250n^2 + 250n - 15000 = 0

Dividing both sides by 250:

n^2 + n - 60 = 0

This equation can be factored as:

(n + 6)(n - 10) = 0

Since we're looking for a positive value for n, the answer is n = 10.

So it will take the jogger 10 days to reach a distance of 15km in training.

(ii) We can use the formula for the sum of a geometric series:

S = a(1 - r^n)/(1 - r)

where S is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.

In this case, we know that the second term is -3, so a = 500 * (-3) = -1500.

We also know that the sum to infinity is 25/2, so S = 25/2.

Plugging in these values and solving for r:

25/2 = (-1500)(1 - r^10)/(1 - r)

r = 1/2 or -1 (discarded since a negative ratio would make the terms alternate in sign, and the second term is negative)

So the common ratio is 1/2.

The total distance the jogger would have run in the training is the sum of the terms of the geometric series:

S = a/(1 - r) = (-1500)/(1 - 1/2) = 3000 meters.

(b)

We know that the second term of the GP is -3, so we can write the first two terms as:

a, ar

where ar = -3.

We also know that the sum to infinity is 25/2, so we can use the formula:

S = a/(1 - r)

25/2 = a/(1 - r)

a = 25/2 - 25r/2

Substituting this value of a into the equation ar = -3:

(25/2 - 25r/2)r = -3

Simplifying:

25r^2 - 50r + 6 = 0

We can solve for r using the quadratic formula:

r = (50

Given that x =  \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\) and y= \(\begin{pmatrix} -9 \\ 15 \end{pmatrix}\) calculate, correct to the nearest degree, the angle between the vectors

Antwortdetails

x =  (−43) $\left(\begin{array}{c}-4\\ 3\end{array}\right)$ and y= (−915) $\left(\begin{array}{c}-9\\ 15\end{array}\right)$

Changing x and y to the form xi+yj

we have:

x= -4i + 3j and y = -9i - 15j

using:

cosØ = xy|x|ly| $\frac{𝑥𝑦}{|𝑥|𝑙𝑦|}$

where Ø is the angle between x and y

xy = (-4i+3j) (-9i - 15j)

-36 + 60 x 0 - 27 x 0- 45 = -81

|x| = √(42 ${}^2$ +32 ${}^2$) =√(16 +9) = √25 = 5

|y| = √(-9)2 ${}^2$ + (-15)2 ${}^2$ = √(81 +225) = √306

cosØ =−815√306 $\frac{-81}{5\surd 306}$

= −9√306170 $\frac{-9\surd 306}{170}$

17.49∗9170 $\frac{17.49\ast 9}{170}$

Ø = cos−1 ${}^{-1}$(-0.1029);

Ø = 95.91°

Ø = 96°

The position vectors of P, Q and R with respect to the origin are (4i-5j), (i+3j) and (-5i+2j) respectively. If PQRM is a parallelogram, find:

(a) the coordinates of M;

(b) the acute angle between \(\overline{PM}\) and \(\overline{PQ}\), correct to the nearest degree.