Question

DU
W
tyujton of the line RS.
(4 marks)
4
Express 0.84554... as a friction in its simplest form,
(04 marks)
5. The coordinates of points A and B are ( 5, 3) and (1,9) respectively,
Find the
a) Mid point of AB
b) Length of AB
6.
1,The function is defined as tx -3x - 2x, Determine the range if the domain is (0,1,2,3)
(04 marks)

2, An open cylinder has a height of 15 cm and a radius of 7cm, calculate the surface area of the cylinder
(04 marks)

3. Given that a = (
and b = 3a, find la + bl.
(04 marks)

9. Apili has shs 20,000,000 on her fixed deposit account in a bank. The bank gives a compound interest
at a rate of 1% per annum Calculate the amount Apili will receive alter 2 years​

Answer 1

Answer:

See Explanation

Explanation:

Solving (4): 0.84554 as a fraction

The above number is to 5 decimal places, so the fraction equivalent is:

$Fraction = \frac{84554}{100000}$

Divide numerator and denominator by 2

$Fraction = \frac{84554/2}{100000/2}$

$Fraction = \frac{42277}{50000}$

Solving (5):

$A = (5,3)$    $B = (1,9)$

The midpoint, M is calculated as follows:

$M = (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Where

$(x_1,y_1) = (5,3)$      $(x_2,y_2) = (1,9)$

$M = (\frac{5+1}{2},\frac{3+9}{2})$

$M = (\frac{6}{2},\frac{12}{2})$

$M = (3,6)$

The distance, D is calculated using:

$D = \sqrt{(x_1 - x_2)^2+(y_1 - y_2)^2}$

$D = \sqrt{(5 - 1)^2+(3 - 9)^2}$

$D = \sqrt{(4)^2+(-6)^2}$

$D = \sqrt{16+36}$

$D = \sqrt{52}$

$D = 7.21$

Solving (1): The expression for f(x) is not clear. So, I'll make use of:

$f(x) = 3x^2 -2x$

$Domain = \{0,1,2,3\}$

Substitute each value of the domain in $f(x) = 3x^2 -2x$

$f(0) = 3(0)^2 - 2(0) = 0 - 0 = 0$

$f(1) = 3(1)^2 - 2(1) = 3 - 2 = 1$

$f(2) = 3(2)^2 - 2(2) = 12 - 4 = 8$

$f(3) = 3(3)^2 - 2(3) = 27 - 6 = 21$

The range is:

$Range = \{0,1,8,21\}$

Solving (2):

$Height (H) = 15cm$    $Radius(R) = 7cm$

Shape: Open Cylinder

The surface area is calculated as:

$Area = 2\pi rh+ \pi r^2$

$Area = 2 * 3.14 * 7 * 15 + 3.14 * 7^2$

$Area = 659.4 + 153.86$

$Area = 813.26$

Solving (3):  

The value of a is not clear. So, I'll assume that a is a

Given that

$b = 3a$

Find $|a+b|$

Substitute 3a for b

$|a + b| = |a + 3a|$

$|a + b| = |4a|$

Absolute value of 4a is 4a. So,

$|a + b| = 4a$

Solving (9):

Given

$P= 20000000$ --- Principal

$R = 1\%\ per\ annum$ --- Rate

$n = 2$ --- Time in years

The amount (A) is calculated as follows:

$A = P* (1 + \frac{R}{100})^n$

$A = 20000000 * (1 + \frac{1}{100})\²$

$A = 20000000 * (101/100)^2$

$A = 20000000 * 1.0201$

$A = 20402000$

Hence, the amount at the end of 2 years is:

$Amount = 20402000$