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3 years ago

(9x) (4x^3) (2x^4y) =

Mathematics

1 answer:

lesya692 [45]3 years ago

4 0

$(9x) (4x^3) (2x^4y) =(9*4*2) x^{1+3+4} y=72 x^{8} y$

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Which terms and 45p^4q have a GCF of 9p^3?

vredina [299]

Answer:

The answer is 18p^3r and 63p^3

Step-by-step explanation:

G.C.F of 18p^3 r and 45p^4q is = 9p^3

18p^3r = 2*3*3*p*p*p*r

45p^4q = 3*3*5*p*p*p*q

Thus the G.C.F is 3*3*p*p*p = 9p^3

G.C.F of 63p^3 and 45p^4q is = 9p^3

63p^3 = 3*3*7*p*p*p

45p^4q = 3*3*5*p*p*p*q

Thus the G.C.F is 3*3*p*p*p = 9p^3

Therefore the answer is 18p^3r and 63p^3....

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3 years ago

Can you guys help me solve this problem

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2 years ago

What multiplication expression can you use to solve 2 divided by 8 4/5?

valina [46]

2*(5/44)

Dividing fractions means take the reciprocal of the second fraction then multiply
Find the reciprocal of 8 4/5
44/5 becomes 5/44
Then multiply by 2

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3 years ago

Please help <br> Thank you

Kaylis [27]

Answer:

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Step-by-step explanation:

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3 years ago

(cot^2x - 1)/(csc^2x) = cos2x

Alex787 [66]

Answer:

Step-by-step explanation:

The idea here is to get the left side simplified down so it is the same as the right side. Consequently, there are 3 identities for cos(2x):

$cos(2x)=cos^2x-sin^2x$ ,

$cos(2x)=1-2sin^2x$ , and

$cos(2x)=2cos^2x-1$

We begin by rewriting the left side in terms of sin and cos, since all the identities deal with sines and cosines and no cotangents or cosecants. Rewriting gives you:

$\frac{\frac{cos^2x}{sin^2x} -\frac{sin^2x}{sin^2x} }{\frac{1}{sin^2x} }$

Notice I also wrote the 1 in terms of sin^2(x).

Now we will put the numerator of the bigger fraction over the common denominator:

$\frac{\frac{cos^2x-sin^2x}{sin^2x} }{\frac{1}{sin^2x} }$

The rule is bring up the lower fraction and flip it to multiply, so that will give us:

$\frac{cos^2x-sin^2x}{sin^2x} *\frac{sin^2x}{1}$

And canceling out the sin^2 x leaves us with just

$cos^2x-sin^2x$ which is one of our identities.

5 0

3 years ago