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enot [183]
3 years ago
14

Jennifer wrote 8 poems at school and Il poems at home. She

Mathematics
2 answers:
swat323 years ago
5 0

Answer:

Nell wrote 4 poems.

Step-by-step explanation:

8+2=10

10-6=4

4= Number of poems that Nell wrote

Paladinen [302]3 years ago
5 0
I’m pretty positive the answer 4
You might be interested in
I really don't get these answers but if you could help me that'll mean a lot thank you for answering my question if you did
mixer [17]
Use KMF (Keep multiply flip) 
keep the first fraction as it is
change sign to multiplication
flip numerator and denominator of last fraction
then finish
6 0
3 years ago
What is the gcf of 32 and 56
Archy [21]

Answer:

The GCF of 32 and 56 are 8.

Step-by-step explanation:

Answer above! Enjoy! :)

8 0
3 years ago
G(x)= 3x − 3; Find g(−6)
Irina18 [472]
<span>g(x)= 3x − 3
</span><span>g(−6)

g(-6) = 3(-6) - 3
g(-6) = -18 - 3
g(-6) = -21

The answer is -21.</span>
4 0
3 years ago
Read 2 more answers
Please help me out :)
xz_007 [3.2K]

Answer: <em>x</em>=12

Explanation:

In a parallelogram, adjacent angles equals 180 (consecutive angles). Therefore, (132-x)+(6x-12)=180

Simplify:

-x+6x+132-12=180

5x+120=180

5x=60

x=12

4 0
3 years ago
From a window 20 feet above the ground, the angle of elevation to the top of a building across
Nikitich [7]

Answer: The answer is 381.85 feet.

Step-by-step explanation:  Given that a window is 20 feet above the ground. From there, the angle of elevation to the top of a building across  the street is 78°, and the angle of depression to the base of the same building is 15°. We are to calculate the height of the building across the street.

This situation is framed very nicely in the attached figure, where

BG = 20 feet, ∠AWB = 78°, ∠WAB = WBG = 15° and AH = height of the bulding across the street = ?

From the right-angled triangle WGB, we have

\dfrac{WG}{WB}=\tan 15^\circ\\\\\\\Rightarrow \dfrac{20}{b}=\tan 15^\circ\\\\\\\Rightarrow b=\dfrac{20}{\tan 15^\circ},

and from the right-angled triangle WAB, we have'

\dfrac{AB}{WB}=\tan 78^\circ\\\\\\\Rightarrow \dfrac{h}{b}=\tan 15^\circ\\\\\\\Rightarrow h=\tan 78^\circ\times\dfrac{20}{\tan 15^\circ}\\\\\\\Rightarrow h=361.85.

Therefore, AH = AB + BH = h + GB = 361.85+20 = 381.85 feet.

Thus, the height of the building across the street is 381.85 feet.

8 0
3 years ago
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