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REY [17]
3 years ago
13

How do I solve an equation like this, one with a negative in front of parentheses?

Mathematics
1 answer:
azamat3 years ago
3 0

The factor outside the parentheses is -1. Distribute using that factor.

2 - (4 - <em>x</em>) = 7<em>x</em> - 5<em>x</em>

<em />

-1 * 4 = -4 and -1 * -<em>x</em> = <em>x</em>

<em />

2 - 4 + <em>x</em> = 7<em>x</em> - 5<em>x</em>

Simplify.

-2 + <em>x</em> = 2<em>x</em>

<em>x</em> = 2<em>x</em> + 2

-<em>x</em> = 2

<em>x</em> = -2

<h3>Answer:</h3>

<em>x</em> = -2

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3 years ago
WILL GIVE BRAINLIEST!!
VMariaS [17]

Answer:

A(-1,4)\to A''(-1,-1)

B(0,2)\to B''(1,0)

C(1,2)\to C''(1,1)

D(-2,4)\to D''(-1,2)

Step-by-step explanation:

The given trapezoid has vertices at A(−1,4), B(0,2), C(1,2) and D(2,4).

The transformation rule for 90° counterclockwise rotation is

(x,y)\to(-y,x)

This implies that:

A(-1,4)\to A'(-4,-1)

B(0,2)\to B'(-2,0)

C(1,2)\to C'(-2,1)

D(2,4)\to D'(-4,2)

This is followed by a translation 3 units to the right.

This also has the rule: (x,y)\to (x+3,y)

A'(-4,-1)\to A''(-1,-1)

B'(-2,0)\to B''(1,0)

C'(-2,1)\to C''(1,1)

D'(-4,2)\to D''(-1,2)

Therefore:

A(-1,4)\to A''(-1,-1)

B(0,2)\to B''(1,0)

C(1,2)\to C''(1,1)

D(-2,4)\to D''(-1,2)

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3 years ago
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IRISSAK [1]
I think the answer is C
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3 years ago
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