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timama [110]
2 years ago
5

Given the following formula, solve for I. P = 2(1 + 6) PLEASE HELP!!!

Mathematics
1 answer:
Pepsi [2]2 years ago
5 0

Step-by-step explanation

Here ...I'm not too sure about the question as you have written

P = 2 ( 1 + 6 ) and ask to solve for L ...

.So I guess the question you have written is wrong .. ...and I guess there should be L in the place of 1.

...So let me correct the question....

P = 2 ( l + 6 )

P = 2l + 12

P - 2l = 12

- 2l = 12 - P

- 2l = - P + 12

Change the sign of both side of the equation...

2l = P + 12

divide both side by the same number...

L = ( P + 12 ) ÷ 2

l =  \frac{1}{2} p - 12 \times  \frac{1}{2}  \\

Calculate the product of rational numbers...

l =  \frac{1}{2} p - 6 \\

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Which of the following equations is equivalent to -2(2 − 2x) = 4 − 8(1 − x) ?
Strike441 [17]

Answer:

0 = 4x

Step-by-step explanation:

-2(2 - 2x) = 4 - 8(1 - x)

-4 + 4x = 4 - 8 + 8x

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8 0
2 years ago
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Percy solved the equation x2 + 7x + 12 = 12. His work is shown below. Is Percy correct? Explain.
antoniya [11.8K]
= > x² + 7x + 12 = 12

= > x² + ( 4 + 3 )x + 12 = 12

= > x² + 4x + 3x + 12 = 12

= > x( x + 4 ) + 3( x + 4 ) = 12

= > ( x + 4 ) ( x + 3 ) = 12


Percy did correct till this step. But by doing like this, Percy can't get the values of the variable x.



Percy should follow the following steps :

= > x² + 7x + 12 = 12


Add -12 on both sides,


= > x² + 7x + 12 - 12 = 12 - 12

= > x² + 7x = 0

= > x( x + 7 ) = 0

= > ( x = 0 ) or ( x + 7 = 0 )

= > ( x = 0 ) or ( x = - 7 )



Hence, required value(s) of x is 0 or -7
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Select the graph for the solution of the open sentence. Click until the correct graph appears. |x| > 3/2
Soloha48 [4]
ANSWER

See attachment

EXPLANATION

The given inequality is

|x| \: > \: \frac{3}{2}

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x\: > \: \frac{3}{2} \: or \: - x\: > \: \frac{3}{2}

Multiply both sides of the second inequality by -1 and reverse the inequality sign.

x\: > \: \frac{3}{2} \: or \: x\: < \: - \frac{3}{2}

The graphical solution to this inequality is shown in the attachment.

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nika2105 [10]
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5 0
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