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Naya [18.7K]
3 years ago
14

What is the area of the figure below? The figure shows a parallelogram. the left side equals 5 inches. The bottom side equals 12

inches. A dashed segment from the top left vertex perpendicular to the bottom left side equals 3 inches.
Mathematics
1 answer:
Strike441 [17]3 years ago
5 0

Answer:

A = 36 in.²

Step-by-step explanation:

The area of a parallelogram can be determined by the formula:

A = bh

The base length is 12 inches and the height is 3 inches.

A = 12in. * 3in.

A = 36 in.²

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Serjik [45]

Answer:

D. 34

Step-by-step explanation:

15+12=27

v-7=27

+7. +7

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v=34

3 0
3 years ago
PLEASE HELP ME THANK YOU IF U DO
Andrews [41]

Answer:

B (r= 1/6s)

Step-by-step explanation:

If you just work out each option, you'll see that "s" multiplied by 1/6 gets r.

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3 years ago
45 +2.50m = 3.75m<br> what is the answer
zhannawk [14.2K]

m = 36

<u>Work</u>

<em>45 + 2.50m = 3.75m</em>

<em>        -3.75m    -3.75m</em>

<em>45 - 1.25m = 0</em>

<em>- 45             - 45</em>

<u><em>-1.25</em></u><em>m = </em><u><em>-45</em></u>

<em>-1.25     -1.25</em>

<em>     </em>m = 36

3 0
3 years ago
Expand using the properties and rules for logarithms
malfutka [58]

Consider expression \log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right).

1. Use property

\log_a\dfrac{b}{c}=\log_ab-\log_ac.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2.

2. Use property

\log_abc=\log_ab+\log_ac.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2.

3. Use property

\log_ab^k=k\log_ab.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2.

4. Use property

\log_{a^k}b=\dfrac{1}{k}\log_ab.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{2^{-1}}2=\\ \\=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+\log_22=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+1.

Answer: correct option is B.

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Answer:

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Step-by-step explanation:

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