All categories

7 months ago

I believe you need to use the quadratic formula for this one, but I am not sure!

Mathematics

1 answer:

Nataly [62]7 months ago

6 0

$4x^2+13=49$

To solve the value of x in the above equation, here are the steps.

1. Subtract 13 on both sides of the equation.

$\begin{gathered} 4x^2+13-13=49-13 \\ 4x^2=36 \end{gathered}$

2. Divide both sides of the equation by 4.

$\begin{gathered} \frac{4x^2}{4}=\frac{36}{4} \\ x^2=9 \end{gathered}$

3. Get the square root on both sides of the equation.

$\begin{gathered} \sqrt[]{x^2}=\sqrt[]{9} \\ x=\pm3 \end{gathered}$

The possible values of x are +3 and -3. The smaller value is -3.

You might be interested in

What is the area of the triangle? PLSSS help me

AveGali [126]

Answer:

The area of the triangle is $A=6 \:units^2$ .

Step-by-step explanation:

The area A of a triangle is given by the formula $A=\frac{1}{2} bh$ where b is the base and h is the height of the triangle.

From the graph, we can see that the base is 3 units and the height is 4 units. Therefore, the area of the triangle is

$A=\frac{1}{2} \cdot3\cdot 4=\frac{12}{2}=6 \:units^2$

8 0

2 years ago

One Endpoint is 9,18 and the midpoint is 14,16 what the other end point

Valentin [98]

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 9}}\quad ,&{{ 18}})\quad 
% (c,d)
&({{ x}}\quad ,&{{ y}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)$

$\bf \left( \cfrac{x+9}{2}~,~\cfrac{y+18}{2} \right)=\stackrel{\textit{midpoint}}{(14,16)}\implies 
\begin{cases}
\cfrac{x+9}{2}=14\\\\
x+9=28\\
\boxed{x=19}\\
-------\\
\cfrac{y+18}{2}=16\\\\
y+18=32\\
\boxed{y=14}
\end{cases}$

7 0

2 years ago

Which topic could be shown in a line graph?

forsale [732]

C favorite food is the answer I believe

7 0

2 years ago

Read 2 more answers

Find The Value Of X When 6-2x = 6x-10x+6. The Value Is

Natasha_Volkova [10]

Answer:

x = 0

Step-by-step explanation:

6 - 2x = 6x - 10x + 6

1. Isolate the variable (x)

Add 2x to both sides

6 - 2x = 6x - 10x + 6 -> 6 = 6x - 10x + 2x + 6

Subtract 6 from both sides

6x - 10x + 2x + 6 -> 0 = 6x - 10x + 2x

2. Combine like terms

0 = 6x - 10x + 2x -> 0 = -2x

3. Solve

Divide both sides by -2

0 = x

*Note: 0 divided by anything is 0 so 0/-2 is 0

(unless it is 0/0 which equals 1)

6 0

2 years ago

Perform the indicated row operations, then write the new matrix.

Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

$\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]$

Operations:

$2R_1 + R_2 ->R_2$

$-3R_1 +R_3 ->R_3$

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

Step-by-step explanation:

The first operation:

$2R_1 + R_2 ->R_2$

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by 2

$2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}0&5&7|1\end{array}\right]$

The second operation:

$-3R_1 +R_3 ->R_3$

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by -3

$-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]$

Hence, the new matrix is:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

3 0

2 years ago

Read 2 more answers