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3 years ago

5(x-2)^2-29=-14 what does x equal

Mathematics

1 answer:

eimsori [14]3 years ago

8 0

Answer:

x = $2+\sqrt{3}$

Step-by-step explanation:

5(x-2)^2= 15 ←add 29 on both sides

(x-2)^2= 3 ← divide by 5 on both sides

x-2= $\sqrt{3}$ ← find the sq root of both sides

x= $2+\sqrt{3}$ ← add 2 on both sides

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Is 11.250 greater or less or equal to 11.25

sasho [114]

Answer:

Equal, 250 and 25 are both 1/4

Step-by-step explanation:

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3 years ago

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HELP When 2((Three-fifths x + 2 and three-fourths y minus one-fourth x minus 1 and one-half y + 3)) is simplified, what is the r

lesya692 [45]

Answer:

<h2> StartFraction 7 over 10 EndFraction x + 2 and one-half y + 6</h2>

Step-by-step explanation:

Given the expression $2(\frac{3x}{5}+2\frac{3y}{4}-\frac{x}{4}-1 \frac{1}{2}y+3)$

To simplify the expression, we need to first collect the like terms of the functions in parentheses as shown;

$= 2(\frac{3x}{5}-\frac{x}{4}-1 \frac{1}{2}y+2\frac{3}{4}y+3)\\= 2(\frac{3x}{5}-\frac{x}{4}- \frac{3}{2}y+\frac{11}{4}y+3)\\$

Then we find the LCM of the resulting function

$= 2(\frac{3x}{5}-\frac{x}{4}- \frac{3}{2}y+\frac{11}{4}y+3)\\= 2(\frac{12x-5x}{20} - (\frac{6y-11y}{4})+3)\\= 2(\frac{7x}{20}- (\frac{-5y}{4})+3 )\\= 2(\frac{7x}{20}+ \frac{5y}{4}+3 )\\= \frac{7x}{10} + \frac{5y}{2} +6\\= \frac{7x}{10} + 2\frac{1}{2}y+6\\$

The final expression gives the required answer

7 0

3 years ago

Find the x- and y-intercept of the line. -7/5x -4y = 7 A. x -intercept is -7/4; y-intercept is 5 B. x -intercept is 5; y-interce

DochEvi [55]

Put y=0
-7/5x-0=7
x=7*-5/7=-5
x intercept=-5
put x=0
-0-4y=7
y=-7/4
y intercept =-7/4

4 0

3 years ago

What is the sum and the product of X to the power of 2, and X

melomori [17]

Answer:

The sum is x^2 + x, the product is x^3.

Step-by-step explanation:

Adding x^2 and x leaves no like terms, so no simplification is required.

Multiplying gives x^2*x, which is x*x*x, which is then x^3

3 0

3 years ago

PLEASE HELP (30 POINTS) Solve the rational equation x/3 = x^2/x + 5 , and check for extraneous solutions.

anygoal [31]

Option C: $x=0$ and $x=\frac{5}{2}$ are the solutions.

Explanation:

The equation is $\frac{x}{3} =\frac{x^{2} }{x+5}$

We shall determine the value of x, by simplifying the equation.

$\begin{aligned} x(x+5) &=3 x^{2} \\ x^{2}+5 x &=3 x^{2} \\ 2 x^{2}-5 x &=0 \\ x(2 x-5) &=0 \end{aligned}$

Thus, $x=0$ and $x=\frac{5}{2}$ are the solutions.

Now, let us check whether the solutions are extraneous solutions.

Let us substitute $x=0$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}&\frac{0}{3}=\frac{0^{2}}{0+5}\\&0=\frac{0}{5}\\&0=0\end{aligned}$

Thus, both sides of the equation are equal.

Hence $x=0$ is a true solution.

Now, Let us substitute $x=\frac{5}{2}$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}\frac{\left(\frac{5}{2}\right)}{3} &=\frac{\left(\frac{5}{2}\right)^{2}}{\left(\frac{5}{2}\right)+5} \\\frac{5}{6} &=\frac{\left(\frac{25}{4}\right)}{\left(\frac{15}{2}\right)} \\\frac{5}{6} &=\frac{5}{6}\end{aligned}$

Thus, both sides of the equation are equal.

Hence, $x=\frac{5}{2}$ is a true solution.

Thus, solutions are not extraneous.

Hence, Option C is the correct answer.

8 0

3 years ago