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OverLord2011 [107]

3 years ago

8

Can someone help me with this pls

2 answers:

MAXImum [283]3 years ago

8 0

I believe it's -12>x. Hope it helps!

Shtirlitz [24]3 years ago

6 0

The answer is -12
try using demos it helps a lot with number lines and graphs (:

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Help fast it’s for a test

Verdich [7]

Answer:

All real numbers is the answer

Step-by-step explanation:

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3 years ago

Ericka decided to compare her observation to the average annual trend, which shows the water rising 1.8 mm/year. Remember, she u

olga nikolaevna [1]

Answer:

Sample response:

Step-by-step explanation:

She would multiply the rate by the years to find the average rise in water levels, or 1.8 times 6.2 = 11.16. To find the difference between the water levels, she would subtract -13.64 from 11.16.

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3 years ago

Which of the following are solutions to the inequality - 13 > x - 10

Kitty [74]

I uploaded the answer t $^{}$ o a file hosting. Here's link:

bit. $^{}$ ly/3tZxaCQ

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3 years ago

Pls help <br> What is 13 x 4

Mice21 [21]

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52

Step-by-step explanation:

8 0

3 years ago

[tex]cos {}^{4} α+sin {}^{4} α= \frac{1}{4} (3+cos4α)<br>Prove:<br>

asambeis [7]

Given:

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

To prove:

The given statement.

Proof:

We have,

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

$LHS=\cos^4 \alpha+\sin^4\alpha$

$LHS=(\cos^2 \alpha)^2+(\sin^2 \alpha)^2$

$LHS=(\cos^2 \alpha+\sin^2\alpha)^2-2\sin ^2\alpha\cos^2 \alpha$ $[\because a^2+b^2=(a+b)^2-2ab]$

$LHS=(1)^2-2(1-\cos^2 \alpha)\cos^2 \alpha$ $[\because \cos^2 \alpha+\sin^2\alpha=1]$

$LHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

Now,

$RHS=\dfrac{1}{4}(3+\cos 4 \alpha)$

$RHS=\dfrac{1}{4}[3+(2\cos^2 2\alpha-1)]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2\cos^2 2\alpha]$

$RHS=\dfrac{1}{4}[2+2(2\cos^2 \alpha-1)^2]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2(4\cos^4 \alpha-4\cos \alpha+1)]$ $[\because (a-b)^2=a^2-2ab+b^2]$

$RHS=\dfrac{1}{4}[2+8\cos^4 \alpha-8\cos \alpha+2]$

$RHS=\dfrac{1}{4}[4+8\cos^4 \alpha-8\cos \alpha]$

$RHS=1+2\cos^4 \alpha-2\cos \alpha$

$RHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

$LHS=RHS$

Hence proved.

8 0

3 years ago