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xxMikexx [17]
4 years ago
10

Write as a numeral 1,000+100+20+1​

Mathematics
2 answers:
Rashid [163]4 years ago
7 0

Answer:

1121

Step-by-step explanation:

Add them together

never [62]4 years ago
6 0

Answer:

The answer is 1,121.

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(-5,9) , ( 11 -39) you have too use m= y2-y1over x2 - x1
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m=\dfrac{y_2-y_1}{x_2-x_1}\\
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4 years ago
5.5/1.1<br> A. 500<br> B. 50<br> C. .05 <br> D. .5<br> E. 5<br> F. .005
IRINA_888 [86]
E. 5 is the correct answer
7 0
3 years ago
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What value of x is in the solution set of 4x – 12 5 16 + 8x?
TEA [102]

Answer:

4x-12= 16 +8x

And we can subtract 4x in both sides and we got:

-12 = 16 +4x

Now we can subtract in both sides 16 and we got:

-28 = 4x

And if we divide both sides by 4 we got:

x = -7

And the best solution would be:

-7

Step-by-step explanation:

For this case we assume the following equation:

4x-12= 16 +8x

And we can subtract 4x in both sides and we got:

-12 = 16 +4x

Now we can subtract in both sides 16 and we got:

-28 = 4x

And if we divide both sides by 4 we got:

x = -7

And the best solution would be:

-7

8 0
3 years ago
Use the power reduction formulas to rewrite the expression. (Hint: Your answer should not contain any exponents greater than 1.)
Lapatulllka [165]

Some useful relations and identities:

\tan x=\dfrac{\sin x}{\cos x}

\sin^2x=\dfrac{1-\cos2x}2

\cos^2x=\dfrac{1+\cos2x}2

By the first relation, we have

\tan^2x\sin^3x=\dfrac{\sin^5x}{\cos^2x}=\dfrac{(\sin^2x)^2\sin x}{\cos^2x}

Applying the two latter identities, we get

\dfrac{\left(\frac{1-\cos2x}2\right)^2\sin x}{\frac{1+\cos2x}2}=\dfrac{\frac{1-2\cos2x+\cos^22x}4\sin x}{\frac{1+\cos2x}2}=\dfrac{(1-2\cos2x+\cos^22x)\sin x}{2(1+\cos2x)}

We can apply the third identity again:

\dfrac{(1-2\cos2x+\cos^22x)\sin x}{2(1+\cos2x)}=\dfrac{\left(1-2\cos2x+\frac{1+\cos4x}2\right)\sin x}{2(1+\cos2x)}=\dfrac{(3-4\cos2x+\cos4x)\sin x}{4(1+\cos2x)}

and this is probably as far as you have to go, but by no means is it the only possible solution.

8 0
3 years ago
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igor_vitrenko [27]
The mean will shift to left
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