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aksik [14]
3 years ago
7

I'm not sure how to answer b.

Mathematics
1 answer:
Brrunno [24]3 years ago
5 0

If you want to check whether a cube root is correct, simply cube your solution and see if you get back the number under the radical. In this case, we can check that \sqrt[3]{-512} =-8 by cubing -8, getting us

(-8)^3=(-8)\cdot(-8)\cdot(-8)=64\cdot(-8)=-512

Our check checks out!

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Translate the following phrase into an expression:The difference between 20 and twice of a number twice of a number
Vitek1552 [10]

Answer: probably D

In math difference means subtraction

Step-by-step explanation:

7 0
3 years ago
Which equation correctly applies the distributive property?
Elodia [21]

Answer:

−3⋅(6.48)=(−3⋅6)+(−3⋅0.4)+(−3⋅0.08)

Step-by-step explanation:

7 0
2 years ago
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Answer the question in the picture
nekit [7.7K]

Recall the angle sum identities:

\sin(x+y)=\sin x\cos y+\cos x\sin y

\cos(x+y)=\cos x\cos y-\sin x\sin y

Now,

\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}

Divide through numerator and denominator by \cos x\cos y to get

\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}

Next, we use the fact that x,y lie in the first quadrant to determine that

\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2

\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}

So we then have

\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}

\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1

Finally,

\tan(x+y)=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3\approx3.73

4 0
3 years ago
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4 0
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How many 5/8 inch binder clips, laid side by side, make a length of 11 1/4 inches?
ElenaW [278]

Answer:

18

Step-by-step explanation:

make 11 1/4 into improper fraction > 45/4

multiply by 2 so it has same denominator > 90/8

90/8 divided by 5/8 = 18

3 0
3 years ago
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