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3 years ago

The perimeter of the semicircle is 25.71 cm. Calculate the value of radius x. Take pi to be 3.142

Mathematics

1 answer:

yan [13]3 years ago

5 0

Answer:

5 cm

Step-by-step explanation:

$Perimeter of semicircle = \frac{1}{2} \times 2\pi \: x +2x\\ 25.71 = 3.14 2\times x+ 2x\\ 25.71 = 5.14 2 x\\ x = \frac{25.71}{5.142} \\ x = 5 \: cm$

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2 years ago

Find the values of x, y, and z. The diagram is not to scale.

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we know that

The sum of the internal angles in the triangle must be $180$ degrees

see the attached figure with letters to better understand the problem

Step $1$

Find the measure of the angle x

In the triangle ABC

$63\°+36\°+x\°=180\°$

solve for x

$99\°+x\°=180\°$

$x\°=180\°-99\°$

$x=81\°$

therefore

the answer Part a) is

the measure of angle x is $81\°$

Step $2$

Find the measure of the angle z

we know that

$x+z=180\°$ --------> by supplementary angles

substitute the value of x

$81\°+z=180\°$

$z=180\°-81\°$

$z=99\°$

therefore

the answer Part b) is

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Step $3$

Find the measure of the angle y

In the triangle ACD

$13\°+z\°+y\°=180\°$

solve for y

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5 0

2 years ago

Read 2 more answers

Match each interval with its corresponding average rate of change for q(x) = (x + 3)2. 1. -6 ≤ x ≤ -4 1 2. -3 ≤ x ≤ 0 -4 3. -6 ≤

MrMuchimi

The average rate of change of a function f(x) in an interval, a < x < b is given by
$\frac{f(b) - f(a)}{b - a}$

Given q(x) = (x + 3)^2

1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by $\frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4$

2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by $\frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3$

3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3$

4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by $\frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1$

5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1$

6.) The average rate of change of q(x) in the interval -6 ≤ x ≤ 0 is given by $\frac{q(0)-q(-6)}{0-(-6)} = \frac{(0+3)^2-(-6+3)^2}{0+6} = \frac{9-9}{6} = \frac{0}{6} =0$

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2 years ago

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