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3 years ago

Can someone please solve this x + 7 = 12

Mathematics

2 answers:

Yuliya22 [10]3 years ago

8 0

Answer:

x=5

Step-by-step explanation:

change to x=12-7

subtract

x=5

lbvjy [14]3 years ago

8 0

Subtract 7 on both sides:
x+7=12
-7 -7
Subtract 12-7:
x=5

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JL = 28, KL= 5x + 6, and JK = 3x + 6,

Alexeev081 [22]

Answer:

KL = 16

Step-by-step explanation:

For this problem, JL is the sum of KL and JK. So we can say this:

JK + KL = JL

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3 years ago

Write the expressions that represents 7 subtracted from 8n

Pani-rosa [81]

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Given:

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

To prove:

The given statement.

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We have,

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

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

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

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

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

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

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

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

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

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

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

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

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Hence proved.

8 0

3 years ago