Is the dilation above an enlargement or a reduction

Mathematics

1 answer:

alexgriva [62]2 years ago

5 0

Reduction...........

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Solve:........................<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B3%7D%20%20%2B%204%20%3D%20%20%5Cfrac%7B4

lakkis [162]

Answer:

$\frac{x}{3} + 4 = \frac{4x - 1}{5} \\ \\ \frac{x}{3} + \frac{12}{3} = \frac{4x - 1}{5} \\ \\ \frac{x + 12}{3} = \frac{4x - 1}{5} \\ \\ 3(4x - 1) = 5(x + 12) \\ \\ 12x - 3 = 5x + 60 \\ \\ 12x - 5x = 60 + 3 \\ \\ 7x = 63 \\ \\ x = \frac{63}{7} \\ \\ x = 9$

7 0

2 years ago

What is the difference?

Crazy boy [7]

Answer:

<h2>D. $\frac{x^{2}+3x-12 }{(x-5)(x+3)(x+7)} \\$ or StartFraction x squared + 3 x minus 12 Over (x + 3) (x minus 5) (x + 7) EndFraction</h2>

Step-by-step explanation:

Given the expression $\frac{x}{x^{2}-2x-15 } - \frac{4}{x^{2} + 2x - 35 }$ , the dfference is expressed as follows;

Step1: First we need to factorize the denominator of each function.

$\frac{x}{x^{2}-2x-15 } - \frac{4}{x^{2} + 2x - 35 }\\= \frac{x}{x^{2}-5x+3x-15 } - \frac{4}{x^{2} + 7x-5x - 35 }\\= \frac{x}{x(x-5)+3(x-5) } - \frac{4}{x( x+ 7)x-5(x +7) }\\= \frac{x}{(x-5)(x+3) } - \frac{4}{(x-5)(x +7) }\\\\$

Step 2: We will find the LCM of the resulting expression

$= \frac{x}{(x-5)(x+3) } - \frac{4}{(x-5)(x +7) }\\= \frac{x(x+7)-4(x+3)}{(x-5)(x+3)(x+7)} \\= \frac{x^{2}+7x-4x-12 }{(x-5)(x+3)(x+7)} \\= \frac{x^{2}+3x-12 }{(x-5)(x+3)(x+7)} \\$