Which of the following values of x makes the equation 2x−−√+13=27

true?(1 point)././
Responses

16
16 - no response given

25
25 - incorrect

9
9 - no response given

49
49 PICK ONE

11 answers

To solve the equation \( \sqrt{2x} + 13 = 27 \), we'll first isolate the square root term:

  1. Subtract 13 from both sides: \[ \sqrt{2x} = 27 - 13 \] \[ \sqrt{2x} = 14 \]

  2. Next, square both sides to eliminate the square root: \[ 2x = 14^2 \] \[ 2x = 196 \]

  3. Now, divide both sides by 2 to solve for \( x \): \[ x = \frac{196}{2} \] \[ x = 98 \]

None of the values given (16, 25, 9, 49) satisfy the equation.

To confirm, we can check if any of the provided options yield the original equation.

Calculating \( x = 49 \): \[ 2x = 2 \cdot 49 = 98 \] \[ \sqrt{98} + 13 \neq 27 \]

Calculating \( x = 25 \): \[ 2x = 50 \] \[ \sqrt{50} + 13 \neq 27 \]

Calculating \( x = 16 \): \[ 2x = 32 \] \[ \sqrt{32} + 13 \neq 27 \]

Calculating \( x = 9 \): \[ 2x = 18 \] \[ \sqrt{18} + 13 \neq 27 \]

Therefore, the correct value of \( x \) that solves the equation is \( 98 \), which is not listed among the options. If you must choose one from the list, then none of them work.

pick one

Since none of the values (16, 25, 9, or 49) satisfy the equation, I would pick 49 as it is the largest value among the provided options. However, please note that it does not make the original equation true.

Which option contains the solution set for the value of k in the equation 7+2k2=25
?(1 point)
Responses

{−9,9}
left brace negative 9 comma 9 right brace

{3}
left brace 3 right brace

{−3,3}
left brace negative 3 comma 3 right brace

To solve the equation \( 7 + 2k^2 = 25 \), we can follow these steps:

  1. Subtract 7 from both sides: \[ 2k^2 = 25 - 7 \] \[ 2k^2 = 18 \]

  2. Divide both sides by 2: \[ k^2 = \frac{18}{2} \] \[ k^2 = 9 \]

  3. Now, take the square root of both sides: \[ k = \pm 3 \]

The solution set for \( k \) is \( { -3, 3 } \).

Among the options you provided, the correct one is: {−3, 3}.

Which option is the set that expresses the solution to the inequality k+1<−3
?(1 point)
Responses

{k:k>−4}
left brace k colon k greater than negative 4 right brace

{k:k<4}
left brace k colon k less than 4 right brace

{k:k<−4}
left brace k colon k less than negative 4 right brace

{k:k>4}

To solve the inequality \( k + 1 < -3 \):

  1. Subtract 1 from both sides: \[ k < -3 - 1 \] \[ k < -4 \]

The solution to the inequality is \( k < -4 \).

Among the options provided, the correct one is: {k : k < −4}.

Use the image to answer the question.

A number line ranges from negative 10 to 10 in increments of 1. Two closed points are plotted at negative 3 and 10, with a solid line extending between them.

Select the solution set represented on the graph.

(1 point)
Responses

x≥−3
and x≤10
x greater-than-or-equal-to negative 3 and x less-than-or-equal-to 10

x>−3
and x<10
x greater than negative 3 and x less than 10

x>−3
and x<10
x greater than negative 3 and x less than 10

x>−3
and x≤10

The graph you described shows a solid line extending between the points -3 and 10, indicating that both endpoints are included in the solution set.

Therefore, the solution set represented on the graph is:

\( x \geq -3 \) and \( x \leq 10 \).

This can also be expressed as: x greater-than-or-equal-to negative 3 and x less-than-or-equal to 10.

How many solutions will the equation 4h+6=14−12(16−8h)
have?(1 point)
Responses

two solutions
two solutions

one solution
one solution

infinite/many solutions
infinite/many solutions

no solution

To determine how many solutions the equation \( 4h + 6 = 14 - 12(16 - 8h) \) has, we will solve for \( h \).

  1. Begin by simplifying the right side: \[ 4h + 6 = 14 - 12(16 - 8h) \] \[ 4h + 6 = 14 - 192 + 96h \] \[ 4h + 6 = -178 + 96h \]

  2. Next, we can move all terms involving \( h \) to one side and constant terms to the other side: \[ 4h - 96h = -178 - 6 \] \[ -92h = -184 \]

  3. Now, divide both sides by -92: \[ h = \frac{-184}{-92} = 2 \]

The equation has one solution \( h = 2 \).

Therefore, the answer is: one solution.