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3 years ago

If BFD and AFC are right angles and

Mathematics

1 answer:

wlad13 [49]3 years ago

4 0

Answer:

Step-by-step explanation:

If BFD is a right angle and it gives you CFD that means that you do 90 - 72 to find CFB cause it 90 all togther

CFB = 18 deg

now it also tells us CFA is a 90 deg angle so we do the same thing

90 - 18 = 72

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Grayson buys milk and oranges at the store.

jonny [76]

Answer: 4.77

Step-by-step explanation:

40.68-2.52=38.16/8=4.77

6 0

2 years ago

Read 2 more answers

Please help me with this problem:)

krok68 [10]

Answer:

(2,1)

Step-by-step explanation:

3x+y = 7

-2x +y = -3

Solve the second equation for y by adding 2x to each side

-2x+2x +y = -3+2x

y = -3 +2x

Substitute into the first equation

3x+ (-3+2x) = 7

Combine like terms

5x -3 = 7

Add 3 to each side

5x-3+3 = 7+3

5x = 10

Divide by 5

5x/5 = 10/5

x=2

Now we need to solve for y

y = -3 +2x

y = -3 +2(2)

y = -3+4

y =1

5 0

4 years ago

Read 2 more answers

Sometimes a dilation is an enlargement, and sometimes it is a reduction. Explain what types of numbers for scale factors causes

Free_Kalibri [48]

If your scale factor has absolute value greater than 1, the dilation is an enlargement.
<span>If your scale factor has abs value less than 1, the dilation is a reduction. </span>
<span>If the scale factor is equal to 1, the image is congruent to the preimage.
</span>
Hope this helps.

4 0

3 years ago

Evaluate 144 to the power of one half

Alborosie

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
-------------------------------\\\\
144^{\frac{1}{2}}\implies \sqrt[2]{144^1}\implies \sqrt{144}\implies 12$

6 0

3 years ago

Read 2 more answers

A suitcase has a lock on it consisting of four numbers. Each number could be any number 0-9. The only restriction is that the la

photoshop1234 [79]

Answer: We have 5000 possible combinations.

Step-by-step explanation:

I guess that we want to find the possible combinations for this lock.

We have 4 numbers, so we have 4 selections.

Now, let's count the number of options in each selection.

in the first selection, we have 10 options (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9)

in the second selection, we have 10 options (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9)

in the third selection, we have 10 options (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9)

in the last selection, we have 5 options (0, 2, 4, 6 and 8)

The total number of combinations is equal to the product of the number of options in each selection:

C = 10*10*10*5 = 5000 combinations.

7 0

3 years ago