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3 years ago

Find the measure of each angle indicated

Mathematics

1 answer:

qaws [65]3 years ago

8 0

Answer:

13. 100* : 14. 111* : 15. 125* : 16. 47*

Step-by-step explanation:

13. Same side interior angles add to 180.

14. Alternate interior angles equal each other.

15. Alternate interior angles equal each other.

16. Opposite angles equal each other.

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- The taxi company charges $0.75 per mile

gladu [14]

Answer: y = .75x + 3

Step-by-step explanation:

y = .75x + 3

a) rate of change is .75x

b) initial value 3

c) independent is .75x

d) dependent is your y

6 0

3 years ago

A trapezoidal prism undergoes a dilation. The surface area of the pre-image is 35 m². The surface area of the image is 315 m². T

Westkost [7]

The original object is called pre-image, and its dilated version is called image. The height of the image for the considered case is 17.4 m

<h3>How does dilation affect length, area, and volume of an object?</h3>

Suppose a figure (pre image) is dilated (dilated image) by scale factor of k.

So, if a side of the figure is of length L units, and that of its similar figure is of M units, then:

L = k × M

where 'k' will be called as scale factor.

The linear things grow linearly like length, height etc.

The quantities which are squares or multiple of linear things twice grow by square of scale factor. Thus, we need to multiply or divide by k² to get each other corresponding quantity from their similar figures' quantities.

So area of first figure = k² × area of second figure

Similarly,

Volume of first figure = k³ × volume of second figure.

It is because we will need to multiply 3 linear quantities to get volume, which results in k getting multiplied 3 times, thus, cubed.

For this case, we're given that;

Surface area of pre-image = 35 m²
Surface area of dilated image = 315 m²
Height of pre-image = 5.8 m
Height of dilated image.

Let the scale fator of dilation be k.

Then:

So area of pre-image = k² × area of dilated image

Thus, we get:

$35 = k^2 \times 315\\\\k = \sqrt{\dfrac{35}{315}} = 1/3$

(took positive root as scaling is done by non-negative factors as only magnitude matters mostly)

Thus, we get:

Height of pre-image = (1/3) × Height of dilated image (say h)

$5.8 = (1/3) \times h\\h = 3 \times 5.8 = 17.4 \rm \: m$

Thus, the original object is called pre-image, and its dilated version is called image. The height of the image for the considered case is 17.4 m

Learn more about dilation here:

brainly.com/question/3266920

4 0

2 years ago

Read 2 more answers

Find the product of z1 and z2, where z1 = 7(cos 40° + i sin 40°) and z2 = 6(cos 145° + i sin 145°).

Oduvanchick [21]

For two complex numbers $z_1=re^{i\theta}=r(\cos\theta+i\sin\theta)$ and $z_2=se^{i\varphi}=s(\cos\varphi+i\sin\varphi)$ , the product is

$z_1z_2=rse^{i(\theta+\varphi)}=rs(\cos(\theta+\varphi)+i\sin(\theta+\varphi))$

That is, you multiply the moduli and add the arguments. You have $z_1=7e^{i40^\circ}$ and $z_2=6e^{i145^\circ}$ , so the product is

$z_1z_2=7\times6(\cos(40^\circ+145^\circ)+i\sin(40^\circ+145^\circ)=42(\cos185^\circ+i\sin185^\circ)=42e^{i185^\circ}$