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3 years ago

PLEASEEE HELPPPP MEE WITH LINE PLOTSSS I DONT UNDERSTAND THISSSS PLEASEE HELPPPP!!!!!!!

Mathematics

1 answer:

Norma-Jean [14]3 years ago

3 0

Answer:

Look at the lightest loaf of bread. It weighs 22 1/3 oz. The heavist loaf weighs 24 1/2. Subtract 24 1/2 and 22 1/3. Then, if the answer to that is 1 1/2 ounces then you agree. If it is not, then disagree. Then, add up all of the weights of the bread and find the answer.

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Which methods correctly solve for the variable s in the equation -2 - 5s =9?

Ivanshal [37]

The answer is C: first add 2 both sides then divide both sides by -5.

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nikitadnepr [17]

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3 years ago

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4 years ago

Which is the completely factored form of 12x^3-60x^2+4x-20

Shkiper50 [21]

Answer:

(12x²+4)(x-5)

Step-by-step explanation:

To factor the expression factor by grouping.

(12x³ - 60x²) + (4x - 20)

12x²(x - 5) 4(x-5)

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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