All categories

3 years ago

So how would you break down the problem(2x-5)^2=49

2 answers:

4 0

First, simplify the equation.

4x2−20x+25=49

Step 2: Subtract 49 from both sides.

4x2−20x+25=49 - 49

4x2−20x−24 =0

Step 3: Factor left side of equation.

4(x+1)(x−6)=0

Step 4: Set factors equal to 0.

x+1=0 or x−6=0

Your answers:
x=−1 or x=6

ivanzaharov [21]3 years ago

3 0

To solve this you first expand the brackets.
4x^2 - 20x + 25 = 49
Then make it equal 0.
4x^2 - 20x - 24 = 0
Now solve the quardratic.

x^2 - 5x - 6 = 0
(x - 6)(x + 1) = 0
x = 6 or x = -1

You might be interested in

Lucy took a taxi from her house to the airport. The taxi company charged a pick-up fee of $3.80 plus $3 per mile. The total fare

prisoha [69]

If you do 39.80/ 3 it would give you 13.2667miles and since the tip was not added that’s when you don’t do anything with that number and 3 because every mile you pay 3$

4 0

3 years ago

Read 2 more answers

Make x the subject! urgent help

Serhud [2]

Answer:

$x = \frac{n}{b - {t}^{2} }$

Step-by-step explanation:

I supposed that is a "n".

Steps as below:

${t}^{2} = b - \frac{n}{x} \\ b - \frac{n}{x} = {t}^{2} \\ - \frac{n}{x} = {t}^{2} - b \\ \frac{n}{x} = - ( {t}^{2} - b) \\ \frac{n}{x} = b - {t}^{2} \\ n = x(b - {t}^{2} ) \\ x(b - {t}^{2} ) = n \\ x = \frac{n}{b - {t}^{2} }$

4 0

1 year ago

Read 2 more answers

How to do the inverse of a 3x3 matrix gaussian elimination.

nata0808 [166]

As an example, let's invert the matrix

$\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}$

We construct the augmented matrix,

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 2 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]$

On this augmented matrix, we perform row operations in such a way as to transform the matrix on the left side into the identity matrix, and the matrix on the right will be the inverse that we want to find.

Now we can carry out Gaussian elimination.

• Eliminate the column 1 entry in row 2.

Combine 2 times row 1 with 3 times row 2 :

2 (-3, 2, 1, 1, 0, 0) + 3 (2, 1, 1, 0, 1, 0)

= (-6, 4, 2, 2, 0, 0) + (6, 3, 3, 0, 3, 0)

= (0, 7, 5, 2, 3, 0)

which changes the augmented matrix to

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]$

• Eliminate the column 1 entry in row 3.

Using the new aug. matrix, combine row 1 and 3 times row 3 :

(-3, 2, 1, 1, 0, 0) + 3 (1, 1, 1, 0, 0, 1)

= (-3, 2, 1, 1, 0, 0) + (3, 3, 3, 0, 0, 3)

= (0, 5, 4, 1, 0, 3)

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 5 & 4 & 1 & 0 & 3 \end{array} \right]$

• Eliminate the column 2 entry in row 3.

Combine -5 times row 2 and 7 times row 3 :

-5 (0, 7, 5, 2, 3, 0) + 7 (0, 5, 4, 1, 0, 3)

= (0, -35, -25, -10, -15, 0) + (0, 35, 28, 7, 0, 21)

= (0, 0, 3, -3, -15, 21)

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 3 & -3 & -15 & 21 \end{array} \right]$

• Multiply row 3 by 1/3 :

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]$

• Eliminate the column 3 entry in row 2.

Combine row 2 and -5 times row 3 :

(0, 7, 5, 2, 3, 0) - 5 (0, 0, 1, -1, -5, 7)

= (0, 7, 5, 2, 3, 0) + (0, 0, -5, 5, 25, -35)

= (0, 7, 0, 7, 28, -35)

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 0 & 7 & 28 & -35 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]$

• Multiply row 2 by 1/7 :

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]$

• Eliminate the column 2 and 3 entries in row 1.

Combine row 1, -2 times row 2, and -1 times row 3 :

(-3, 2, 1, 1, 0, 0) - 2 (0, 1, 0, 1, 4, -5) - (0, 0, 1, -1, -5, 7)

= (-3, 2, 1, 1, 0, 0) + (0, -2, 0, -2, -8, 10) + (0, 0, -1, 1, 5, -7)

= (-3, 0, 0, 0, -3, 3)

$\left[ \begin{array}{ccc|ccc} -3 & 0 & 0 & 0 & -3 & 3 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]$

• Multiply row 1 by -1/3 :

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

So, the inverse of our matrix is

$\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}^{-1} = \begin{bmatrix}0&1&-1\\1&4&-5\\-1&-5&7\end{bmatrix}$

6 0

2 years ago

How much higher is checkpoint 4 than checkpoint 1 checkpoint 1= -86 checkpoint 4= 2,177

Rina8888 [55]

Answer is 2263.
Checkpoint 1 is -86
Checkpoint 4 is 2177
Subtract -86 from 2177....2177-(-86)
=2263
2263 is how much higher checkpoint 1 (-86) is from checkpoint 4 (2177)

Hope this helps! :)

6 0

2 years ago

Read 2 more answers

Choose the point-slope form of the equation below that represents the line that passes through the points (-6, 4) and (2, 0).

Arte-miy333 [17]

Point slope form
y-y1=m(x-x1)
slope=m
a point is (x1,y1)

slope=(y2-y1)/(x2-x1)
given
(-6,4) and (2,0)
slope=(0-4)/(2-(-6))=-4/(2+6)=-4/8=-1/2

a point is (-6,4)
x1=-6
y1=4

y-4=-1/2(x+6)
not listed

5 0

3 years ago