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melomori [17]
2 years ago
15

I NEED HELP ASAP IM ON A TEST ABOUT PROPORTIONAL RELATIONSHIP WILL GIVE POINTS AND BRAINLIST

Mathematics
1 answer:
JulijaS [17]2 years ago
8 0

Answer:

12/1

Step-by-step explanation:

k = y/x, where y and x is how much is added to the respective variables each point.

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SOLVE: 6 and 1/12 divided by 4 and2/5​
Firdavs [7]

Answer:

1 and 101/264

Step-by-step explanation:

:)

(:

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Hope this helps!

-Josh

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(brainliest?)

7 0
3 years ago
Factor completely 12x4 + 6x3 + 18x2.
Semmy [17]

Answer:

B)3x^2 (4*x^2 + 2x + 6)

Step-by-step explanation:

Step 1: Find the Greatest common factor of the given expression.

12x^4 + 6x^3 + 18x^2

The above expression can be written as .

= 2*2*3*x^4 + 2*3*x^3 + 2*3*3*x^2

Here 3x^2 is prime factor


Step 2: Let's take out the 3x^{2} and write the remaining terms in the parenthesis.

= 3x^2 (2*2*x^2 + 2x + 2*3)

= 3x^2 (4x^2 + 2x + 6)

Therefore, the answer is B)3x^2 (4*x^2 + 2x + 6)

Thank you.

8 0
4 years ago
The density d of a substance is given by the formula d=mV, where m is its mass and V is its volume. Solve the formula for m.
NikAS [45]

Answer:m=d/V

Step-by-step explanation:

To solve for m, you must divide V from each side. Leaving you with m=d/V. Hope this helped:)

5 0
4 years ago
Read 2 more answers
Please help me!! How do you factor this???<br> 16x^5+12xy-9y^5
Aleksandr [31]
First simplifying step by step.

<span>16x^5+12xy-9y^5

Answer is = </span><span><span><span>16<span>x^5</span></span>+<span><span>12x</span>y</span></span>+</span>−<span>9<span>y<span>5

It will help you.
</span></span></span>
7 0
3 years ago
Need help on this calculus
N76 [4]
\displaystyle\int\sin^3t\cos^3t\,\mathrm dt

One thing you could do is to expand either a factor of \sin^2t or \cos^2t, then expand the integrand. I'll do the first.

You have

\sin^2t=1-\cos^2t

which means the integral is equivalent to

\displaystyle\int\sin t(1-\cos^2t)\cos^3t\,\mathrm dt

Substitute u=\cos t, so that \mathrm du=-\sin t\,\mathrm dt. This makes it so that the integral above can be rewritten in terms of u as

\displaystyle-\int(1-u^2)u^3\,\mathrm du=\int(u^5-u^3)\,\mathrm du

Now just use the power rule:

\displaystyle\int(u^5-u^3)\,\mathrm du=\dfrac16u^6-\dfrac14u^4+C

Back-substitute to get the antiderivative back in terms of t:

\dfrac16\cos^6t-\dfrac14\cos^4t+C
4 0
3 years ago
Read 2 more answers
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