Solved on Dec 08, 2023

7. Find $m$ such that $\frac{2 x}{m}+\frac{3 a}{x}=\frac{5}{c}$
8. Solve $x^{2}=7-8 x$ by completing the square
9. Solve $189 x=3 x^{3}+6 x^{2}$ by factoring
10. Solve $7=2+\sqrt{x+3}$
11. Solve $3 x^{0}-4 x\left(-1-3^{0}\right)=5(x+3)$

STEP 1

Assumptions for problem 7
1. We are given the equation $\frac{2x}{m} + \frac{3a}{x} = \frac{5}{c}$ .
2. We need to find the value of $m$ .
3. $x$ , $a$ , and $c$ are constants and $x \neq 0$ .

STEP 2

Multiply both sides of the equation by $mcx$ to eliminate the denominators.
$mcx \left( \frac{2x}{m} + \frac{3a}{x} \right) = mcx \left( \frac{5}{c} \right)$

STEP 3

Distribute $mcx$ on both sides of the equation.
$2cx^2 + 3amc = 5mx$

STEP 4

We need to isolate $m$ on one side of the equation. To do this, we can move all terms containing $m$ to one side and the rest to the other side.
$3amc - 5mx = -2cx^2$

STEP 5

Factor out $m$ from the left side of the equation.
$m(3ac - 5x) = -2cx^2$

STEP 6

Divide both sides by $(3ac - 5x)$ to solve for $m$ .
$m = \frac{-2cx^2}{3ac - 5x}$
This is the value of $m$ in terms of $x$ , $a$ , and $c$ .

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7. Find $m$ such that $\frac{2 x}{m}+\frac{3 a}{x}=\frac{5}{c}$
8. Solve $x^{2}=7-8 x$ by completing the square
9. Solve $189 x=3 x^{3}+6 x^{2}$ by factoring
10. Solve $7=2+\sqrt{x+3}$
11. Solve $3 x^{0}-4 x\left(-1-3^{0}\right)=5(x+3)$

STEP 1

Assumptions for problem 7
1. We are given the equation $\frac{2x}{m} + \frac{3a}{x} = \frac{5}{c}$ .
2. We need to find the value of $m$ .
3. $x$ , $a$ , and $c$ are constants and $x \neq 0$ .

STEP 2

Multiply both sides of the equation by $mcx$ to eliminate the denominators.
$mcx \left( \frac{2x}{m} + \frac{3a}{x} \right) = mcx \left( \frac{5}{c} \right)$

STEP 3

Distribute $mcx$ on both sides of the equation.
$2cx^2 + 3amc = 5mx$

STEP 4

We need to isolate $m$ on one side of the equation. To do this, we can move all terms containing $m$ to one side and the rest to the other side.
$3amc - 5mx = -2cx^2$

STEP 5

Factor out $m$ from the left side of the equation.
$m(3ac - 5x) = -2cx^2$

STEP 6

Divide both sides by $(3ac - 5x)$ to solve for $m$ .
$m = \frac{-2cx^2}{3ac - 5x}$
This is the value of $m$ in terms of $x$ , $a$ , and $c$ .

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STEP 1

Assumptions for problem 71. We are given the equation 2xm+3ax=5c\frac{2x}{m} + \frac{3a}{x} = \frac{5}{c}m2x​+x3a​=c5​.2. We need to find the value of mmm.3. xxx, aaa, and ccc are constants and x≠0x \neq 0x=0.

STEP 2

Multiply both sides of the equation by mcxmcxmcx to eliminate the denominators.mcx(2xm+3ax)=mcx(5c)mcx \left( \frac{2x}{m} + \frac{3a}{x} \right) = mcx \left( \frac{5}{c} \right)mcx(m2x​+x3a​)=mcx(c5​)

STEP 3

Distribute mcxmcxmcx on both sides of the equation.2cx2+3amc=5mx2cx^2 + 3amc = 5mx2cx2+3amc=5mx

STEP 4

We need to isolate mmm on one side of the equation. To do this, we can move all terms containing mmm to one side and the rest to the other side.3amc−5mx=−2cx23amc - 5mx = -2cx^23amc−5mx=−2cx2

STEP 5

Factor out mmm from the left side of the equation.m(3ac−5x)=−2cx2m(3ac - 5x) = -2cx^2m(3ac−5x)=−2cx2

STEP 6

Divide both sides by (3ac−5x)(3ac - 5x)(3ac−5x) to solve for mmm.m=−2cx23ac−5xm = \frac{-2cx^2}{3ac - 5x}m=3ac−5x−2cx2​This is the value of mmm in terms of xxx, aaa, and ccc.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Assumptions for problem 7
1. We are given the equation $\frac{2x}{m} + \frac{3a}{x} = \frac{5}{c}$ .
2. We need to find the value of $m$ .
3. $x$ , $a$ , and $c$ are constants and $x \neq 0$ .

Multiply both sides of the equation by $mcx$ to eliminate the denominators.
$mcx \left( \frac{2x}{m} + \frac{3a}{x} \right) = mcx \left( \frac{5}{c} \right)$

Distribute $mcx$ on both sides of the equation.
$2cx^2 + 3amc = 5mx$

We need to isolate $m$ on one side of the equation. To do this, we can move all terms containing $m$ to one side and the rest to the other side.
$3amc - 5mx = -2cx^2$

Factor out $m$ from the left side of the equation.
$m(3ac - 5x) = -2cx^2$

Divide both sides by $(3ac - 5x)$ to solve for $m$ .
$m = \frac{-2cx^2}{3ac - 5x}$
This is the value of $m$ in terms of $x$ , $a$ , and $c$ .