Pls help!! 15 points!

Mathematics

2 answers:

Vesna [10]2 years ago

6 0

When trapezoidal prism undergoes a dilation, the size of the prism changes

The height of the image is 52.2 meters

<h3>How to determine the height of the image?</h3>

The given parameters are:

A1 = 35 m²

A2 = 315 m²

H1 = 5.8 m

To calculate the height of the image, we make use of the following equivalent ratio

A1 : H1 = A2 : H2

So, we have:

35 : 5.8 = 315 : H

Express as fractions

35/5.8 = 315/H

Make H the subject

H = 315 * 5.8/35

Evaluate the product

H = 52.2

Hence, the height of the image is 52.2 meters

You might be interested in

A particle moves along a line with a velocity v(t)=t2−t−6, measured in meters per second. Find the total distance the particle t

ahrayia [7]

Answer:

10.67 m

Step-by-step explanation:

Velocity is given as;

v(t) = t² − t − 6

To get the distance, we will integrate the velocity equation.

x(t) = ∫v(t) = (t³/3) - (t²/2) - 6t

Thus, from t=0 seconds to t=4 seconds., we have;

(t³/3) - (t²/2) - 6t between 0 and 4.

Thus;

Total Distance = [(4³/3) - (4²/2) - 6(4)] - [(0³/3) - (0²/2) - 6(0)]

Total distance = (64/3) - 8 - 24 = -10.67 m

Distance can't be negative and so we take the absolute value which is 10.67 m

7 0

3 years ago

How to solve this :') please help

blsea [12.9K]

Answer:

3/2

Step-by-step explanation:

sin(3π/4 - β) = sin(3π/4)cosβ - cos(3π/4)sinπ =

$\sin(\frac{3\pi}{4}-\beta)=\sin(\frac{3\pi}{4})\cos\beta-\cos\frac{3\pi}{4}\sin\beta\\=\frac{1}{\sqrt{2}}\cos\beta-(-\frac{1}{\sqrt{2}})\sin\beta\\\\=\frac{1}{\sqrt{2}}(\cos\beta +\sin\beta)\\\\\sin^2(\frac{3\pi}{4}-\beta)=\frac{1}{2}\cos\beta +\sin\beta)^2=\frac{1}{2}(\cos^2\beta +\sin^2\beta+2\sin\beta\cos\beta\\=\frac{1}{2}(1+\sin2\beta)=\frac{1}{2}(1-\frac{1}{5}) = \frac{2}{5}\\$