Ask question

Ask question

All categories

3 years ago

10

Follow the process of completing the square to solve 2x2 + 8x - 12 = 0. After adding B2 to both sides of the equation in step 4,

what is the constant on the right side of the equation?

2 answers:

Zepler [3.9K]3 years ago

6 0

Answer:

160

Step-by-step explanation:

Katena32 [7]3 years ago

3 0

2x^2 + 8x - 12 = 0..divide by 2
x^2 + 4x - 6 = 0
x^2 + 4x = 6...add 4 to both sides of the equation
x^2 + 4x + 4 = 6 + 4
(x + 2)^2 = 10....<== ur constant is 10
x + 2 = (+-)sqrt 10
x = -2 (+ - ) sqrt 10

x = -2 + sqrt 10
x = -2 - sqrt 10

You might be interested in

Daniel rollerbladed 3 miles in 12 minutes what was his speed in miles per hour

a_sh-v [17]

Daniel is rollerblading at 15 miles per hour

Daniel can go 3 miles in 12 minutes so we do 12÷3 to find out how many minutes it takes to go 1 mile which is 4 now we do 60÷4 to get 15

Alternatively we could do 60÷12 to get 5 then do 3×5 to get 15

7 0

3 years ago

Read 2 more answers

Before a sale, an item’s price was $28.00. After the sale discount the price became $21.00. What was the percent of the sale dis

NARA [144]

The discount was 25% off use 21/28 which is 75 percent of what it was before. 25 percent off.

4 0

3 years ago

Two collinear points on a line are given in the table below:

katrin [286]

Answer:

$(4,3)$ and $(7,2)$ do not lie on the line

Step-by-step explanation:

Given

$(0,0)\ and\ (2,1)$

Required

Determine which points that are not on the line

First, we need to determine the slope (m) of the line:

$m = \frac{y_2 - y_1}{x_2- x_1}$

Where

$(x_1,y_1) = (0,0)$

$(x_2,y_2) = (2,1)$

So;

$m = \frac{y_2 - y_1}{x_2- x_1}$

$m = \frac{1 - 0}{2-0}$

$m = \frac{1}{2}$

Next, we determine the line equation using:

$y - y_1 = m(x -x_1)$

Where

$m = \frac{1}{2}$

$(x_1,y_1) = (0,0)$

$y - y_1 = m(x -x_1)$ becomes

$y - 0 = \frac{1}{2}(x - 0)$

$y = \frac{1}{2}x$

To determine which point is on the line, we simply plug in the values of x to in the equation check.

For $(4,2)$

$x = 4$ and $y =2$

Substitute 4 for x and 2 for y in $y = \frac{1}{2}x$

$2 = \frac{1}{2} * 4$

$2 = \frac{4}{2}$

$2=2$

<em>This point is on the graph</em>

<em></em>

For $(4,3)$

$x = 4$ and $y = 3$

Substitute 4 for x and 3 for y in $y = \frac{1}{2}x$

$3 = \frac{1}{2} * 4$

$3 = \frac{4}{2}$

$3 \neq 2$

<em>This point is not on the graph</em>

<em></em>

For $(7,2)$

$x = 7$ and $y = 2$

Substitute 7 for x and 2 for y in $y = \frac{1}{2}x$

$2 = \frac{1}{2} * 7$

$2 = \frac{7}{2}$

$2 \neq 3.5$

<em></em>

<em>This point is not on the graph</em>

<em></em>

<em></em> $(\frac{4}{8},\frac{2}{8})$ <em></em>

<em></em> $x = \frac{4}{8}$ and<em> </em> $y = \frac{2}{8}$ <em></em>

<em>Substitute </em> $\frac{4}{8}$ <em> for x and </em> $\frac{2}{8}$ <em> for y in </em> $y = \frac{1}{2}x$ <em></em>

<em></em> $\frac{2}{8} = \frac{1}{2} * \frac{4}{8}$ <em></em>

<em></em> $\frac{2}{8} = \frac{1 * 4}{8 * 2}$ <em></em>

<em></em> $\frac{2}{8} = \frac{4}{16}$ <em></em>

<em></em> $\frac{1}{4} = \frac{1}{4}$ <em></em>

<em></em>

<em>This point is on the graph</em>

3 0

3 years ago

Where decimal go in 288

Andreyy89

Be more specific. The decimal can be anywhere give more detail

6 0

3 years ago

Find area and perimeter

alisha [4.7K]

Answer:

Perimeter

P = P1+P2

P1 = 4a P2=2(a+b)

P1=4*5 P2=2(3+5)

P1=20 P2=2*8

P2=16

P = 20+16

P=36

Area

A=A1+A2

A1= $a^{2}$ A2=a*b

A1= $5^{2}$ A2=3*5

A1=25 A2=15

A=25+15

A=40

5 0

3 years ago