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3 years ago

11

Could you solve it? I don't know how to solve it:(

1 answer:

Nikitich [7]3 years ago

5 0

Answer:

-1, 4

Step-by-step explanation:

$log_2x+ log_2(x-3)=2 \\ log_2 \{x(x-3) \}=2 \\ x(x - 3) = {2}^{2} \\ {x}^{2} - 3x = 4 \\ {x}^{2} - 3x - 4 = 0 \\ {x}^{2} - 4x + x - 4 = 0 \\ x(x - 4) + 1(x - 4) = 0 \\ (x - 4)(x + 1) = 0 \\ x - 4 = 0 \: \: or \: \: x + 1 = 0 \\ x = 4 \: \: or \: \: x = - 1 \\ \huge \red{ \boxed{x = \{ - 1, \: \: 4 \}}}$

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Fill in the gaps in the arithmetic sequence -3, _ ,_ , _, _, _ 12

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Andrew is paid $4 per hour for the first 30 hours he works each week. he makes $5 per hour for each hour he works over 30 hours

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3 years ago

4/5 divided by 1/3 plus 1/5 minus 3/5

r-ruslan [8.4K]

Answer:

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}=-12$

Step-by-step explanation:

Considering the expression

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}$

Solution Steps:

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}$

as

$\mathrm{Combine\:the\:fractions\:}\frac{1}{5}-\frac{3}{5}:\quad -\frac{2}{5}$

so

$=\frac{\frac{4}{5}}{\frac{1}{3}-\frac{2}{5}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}$

$=\frac{4}{5\left(\frac{1}{3}-\frac{2}{5}\right)}$

join $\frac{1}{3}-\frac{2}{5}:\quad -\frac{1}{15}$

so

$=\frac{4}{5\left(-\frac{1}{15}\right)}$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$=\frac{4}{-5\cdot \frac{1}{15}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{-b}=-\frac{a}{b}$

$=-\frac{4}{5\cdot \frac{1}{15}}$

$\mathrm{Multiply\:}5\cdot \frac{1}{15}\::\quad \frac{1}{3}$

so

$=-\frac{4}{\frac{1}{3}}$

$\mathrm{Simplify}\:\frac{4}{\frac{1}{3}}:\quad \frac{12}{1}$

so

$=-\frac{12}{1}$

$\mathrm{Apply\:rule}\:\frac{a}{1}=a$

$=-12$

Therefore

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}=-12$

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3 years ago