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3 years ago

Quadilateral ABCD is similar to quadilateral EFGD Help?

Mathematics

2 answers:

Kay [80]3 years ago

8 0

The figure shown below is an example of ratio and proportion. A ratio is a comparison between two different things. A proportion is two ratios that have been set equal to each other. Note that in the figure shown below, the form and angle of quadilateral ABCD is similar to quadilateral EFGD. So if we have angle ABC, it is proportional to angle EFG or angle E. From the question, angle CGF is proportional to angle GDE or angle D.

Hunter-Best [27]3 years ago

4 0

This is a sadly defective question.

It wants you to mark "Angle D", because Angle-D is
equal to angle-CGF ... they're the same size.

But they are not "corresponding" angles.

Angle-CGF is an angle outside of the smaller figure.
The corresponding angle of the larger figure would be
the angle above and outside of corner-C, if the side DC
were extended a little bit to make an angle up there.

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A. 3 B. 3.1 C. 3.8 D. 11.6

pishuonlain [190]

Mean = (2+4+6+5+2)/5 = 19/5 = 3.8

answer

C. 3.8

4 0

3 years ago

A family has an annual income of $30,600 Of this, 1/4 is spent for food, 1/5 for housing, 1/10 for clothing, 1/9 for savin

Ksenya-84 [330]

Answer:

$6,120

Step-by-step explanation:

1/5 (one-fifth) of the family's annual income ($30,600) is spent on housing- as the word problem states.

You have to find 1/5 (one-fifth) of* $30,600.

Write it out as an equation.

$\frac{1}{5} \times 30600$

*The of in the word equation (1/5 of 30600) simply means multiplication.

4 0

2 years ago

Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)

Mnenie [13.5K]

One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
$(x-h)^2+(y-k)^2=r^2$
the center is (h,k) and the radius is r

and the distance formula is
distance between $(x_1,y_1)$ and $(x_2,y_2)$ is
$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle

so
$(x+2)^2+(y-3)^2=18$
$(x-(-2))^2+(y-3)^2=(\sqrt{18})^2$
$(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2$

the radius is $3\sqrt{2}$
center is (-2,3)

find distance between (8,4) and (-2,3)

$D=\sqrt{(8-(-2))^2+(4-3)^2}$
$D=\sqrt{(8+2)^2+(1)^2}$
$D=\sqrt{10^2+1}$
$D=\sqrt{100+1}$
$D=\sqrt{101}$

$r=3\sqrt{2}$ ≈4.2
$D=\sqrt{101}$ ≈10.04

do r<D

(8,4) is outside the circle

6 0

3 years ago

In order to solve the following system of equations by subtraction, which of the following could you do before subtracting the e

Whitepunk [10]

I can't see the options, but you would need to multiply (4x - 2y = 7) by 3, and (3x - 3y = 15) by 4 so that you get 12x in both equations. Then when you subtract, you eliminate 12x. I hope this helps!

7 0

3 years ago

Read 2 more answers

Giving brainlist to whoever answers

SpyIntel [72]

Answer:

144m2

Step-by-step explanation:

I think very carefully on my brain is it cirrect?

3 0

3 years ago