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3 years ago

Why do you think it is important to be able to rewrite the same expression in different ways?

Mathematics

1 answer:

Alex73 [517]3 years ago

5 0

Answer:

Cause If you write it one way and it is correct then you can write it another way but it is still correct

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Two angles are complimentary. One angle measures 69 degrees. The other measures ( 5n + 1). What is the value of n?

iren [92.7K]

Answer:

Step-by-step explanation:

complimentary angles, when added = 90 degrees

69 + 5n + 1 = 90

5n + 70 = 90

5n = 90 - 70

5n = 20

n = 20/5

n = 4 <====

6 0

4 years ago

What is the nth term of the sequence? 4, 2, 0, –2, ..

tatuchka [14]

' n ' the 'n'th term

1    4
2    2
3    0
4 -2 .

In general, the 'n'th term of the sequence is A(n) = 6 - 2n .

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3 years ago

It cost $120 to buy the materials for crown molding in the completed office. (Crown molding is the decorative border where the w

scoundrel [369]

Answer:75

Step-by-step explanation:

4 0

2 years ago

Any sort of help is appreciated, ty

Semenov [28]

The answer would be 1952 because the display of flag makes a right triangle so we will use Pythagorean theorem in order to find the sides and the sides total is 244 according to theorem, so we multiply by 8 to get the 8 flag display. <span />

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4 years ago

Find the volume of the following solid. The solid between the cylinder f(x,y)equals=e Superscript negative xe−x and the region

PilotLPTM [1.2K]

It is hard to comprehend your question. As far as I understand:

f(x,y) = e^(-x)

Find the volume over region R = {(x,y): 0<=x<=ln(6), -6<=y <= 6}.

That is all I understood. It would be easier to understand with a picture or some kind of visual aid.

Anyways, to find the volume between the surface and your rectangular region R, we must evaluate a double integral of f on the region R.

$\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy$

Now evaluate,

$\int_{0}^{ln(6)}e^{-x}dx$

which evaluates to, 5/6 if I did the math correct. Correct me if I am wrong.

Now integrate this w.r.t. y:

$\int_{-6}^{6}\frac{5}{6}dy = 10$

So,

$\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy = 10$

7 0

3 years ago