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Sergeeva-Olga [200]

3 years ago

10

((-2^2)(-1^-3))•((-2^-3)(-1^5))^-2

1 answer:

Vitek1552 [10]3 years ago

3 0

Answer:

256

Step-by-step explanation:

A calculator works well for this.

_____

None of the minus signs are subject to the exponents (because they are not in parentheses, as (-1)^5, for example. Since there are an even number of them in the product, their product is +1 and they can be ignored.

1 to any power is still 1, so the factors (1^n) can be ignored.

After you ignore all of the things that can be ignored, your problem simplifies to ...

(2^2)(2^-3)^-2

The rules of exponents applicable to this are ...

(a^b)^c = a^(b·c)

(a^b)(a^c) = a^(b+c)

Then your product simplifies to ...

(2^2)(2^((-3)(-2)) = (2^2)(2^6)

= 2^(2+6)

= 2^8 = 256

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mr Goodwill [35]

$2\sin\alpha-\cos\alpha=2$

Consider the substitution $\tan\dfrac\alpha2=\beta$ . Then by the double angle identities we get

$\sin\alpha=2\sin\dfrac\alpha2\cos\dfrac\alpha2$

$\cos\alpha=\cos^2\dfrac\alpha2-\sin^2\dfrac\alpha2$

We also have

$\tan\dfrac\alpha2=\beta\implies\begin{cases}\sin\dfrac\alpha2=\dfrac\beta{\sqrt{1+\beta^2}}\\\\\cos\dfrac\alpha2=\dfrac1{\sqrt{1+\beta^2}}\end{cases}$

so that

$\sin\alpha=\dfrac{2\beta^2}{1+\beta^2}$

$\cos\alpha=\dfrac{1-\beta^2}{1+\beta^2}$

and the original equation has been transformed to

$\dfrac{4\beta^2-(1-\beta^2)}{1+\beta^2}=2$

Solve for $\beta$ :

$5\beta^2-1=2+2\beta^2$

$3\beta^2=3$

$\beta^2=1$

$\beta=\pm1$

Solving for $\alpha$ gives

$\tan\dfrac\alpha2=-1\implies\dfrac\alpha2=-\dfrac\pi4+n\pi\implies\alpha=-\dfrac\pi2+2n\pi$

$\tan\dfrac\alpha2=1\implies\dfrac\alpha2=\dfrac\pi4+n\pi\implies\alpha=\dfrac\pi2+2n\pi$

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$\cos(x+2n\pi)=\cos x$

$\sin(x+2n\pi)=\sin x$

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$\sin\alpha=\sin\left(\pm\dfrac\pi2+2n\pi\right)=\sin\left(\pm\dfrac\pi2\right)=\pm1$

$\cos\alpha=\cos\left(\pm\dfrac\pi2+2n\pi\right)=\cos\left(\pm\dfrac\pi2\right)=0$

and we find that

$\sin\alpha+2\cos\alpha=\pm1$

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3 years ago

A bunch of bananas weighs 8 llbs. Each banana in the bunch weighs exactly 0.5 llbs. How many bananas are in a bunch?

Alenkinab [10]

Answer:

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Step-by-step explanation:

to solve this you should start with the total amount the bananas weigh...

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now you should divide by how much one is to find out how many are in this bunch.

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this comes out to 16

therefore, there are 16 bananas in a bunch.

i hope this helps!!

stay safe!

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3 years ago

Question 1 (1 point)

eimsori [14]

Answer:

Number 1

Step-by-step explanation:

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3 years ago