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2 years ago

HELPP ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

DaniilM [7]2 years ago

7 0

Answer:

a warm front is happening

Nookie1986 [14]2 years ago

6 0

Answer:

hot

Step-by-step explanation:

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The expression cos(A-B) - cos(A+B) is equal to

olchik [2.2K]

Cos (A-B) - cos (A+B)
= (cosAcosB +sinAsinB) - (cosAcosB - sinAsinB)
=2sinAsinB

Ans: 4

6 0

2 years ago

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Put Answer below....

xenn [34]

The value of M equals 70°

3 0

2 years ago

Use the quadratic formula to find the roots: -3x^2 + 27x = 42

ch4aika [34]

Answer:

x=2,x=7

Step-by-step explanation:

-3x^2+27x=42

x^2-9x+14=0

x^2-2x-7x+14=0

x*(xi2)-7(x-2)=0

(x-2)*(x-7)=0

x-2=0

x-7=0

x=2

x=7

8 0

3 years ago

One fourth the sum of x and ten is identical to x minus 4."

Lynna [10]

Answer and Step-by-step explanation:

We are given the word format of an equation.

Let's right it out in numerical form.

$\frac{1}{4} (x + 10) = x - 4$

The first part says One fourth the sum of x and 10

The sum of x and 10 would be shown as x + 10, and one fourth of that means that $\frac{1}{4}$ is multiplying (x + 10).

In this case, identical would mean equal to.

So, $\frac{1}{4}$ (x + 10) would be = to something.

It says that it is identical to x minus 4 (x - 4).

$\frac{1}{4} (x + 10) = x - 4$ is the equation.

#teamtrees #PAW (Plant And Water)

5 0

2 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

2 years ago

Read 2 more answers