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4 years ago

11

Please Help Me Will give 10 points

1 answer:

vazorg [7]4 years ago

8 0

Answer:

I would say 1/2 but this question is really confusing.

Step-by-step explanation:

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What is <br> the equation of the line that passes through the points (-6,6) and (-1,-9)

IgorLugansk [536]

Answer:

The answer to the question provided is y = -3x - 12.

Step-by-step explanation:

❃Incase you forgot what the linear equation formula is ☟

$y = mx + b$

❃Incase you also forgot, what the slope formula is ☟

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

➊ First: We are going to be solving for the slope.

$m = \frac{ - 9 - 6}{ - 1 - ( - 6)} = \frac{ - 15}{ \: \: \: 5} = - 3$

➋Second: We find the y-intercept.

$y = mx + b \\ 6 = - 3( - 6) + b \\ 6 = 18 + b \\ \frac{ - 18 = - 18 \: \: \: \: \: \: \: \: \: \: \: \: }{ - 12 = b}$

➌Third: Plug in.

$y = - 3x - 12$

6 0

3 years ago

Find the value of the value of x.

mojhsa [17]

X=15

Hope this helps I had the same problem. That’s what I got.

3 0

3 years ago

Read 2 more answers

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

<u>case a)</u> $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

<u>case b)</u> $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

<u>case c)</u> $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

<u>case d)</u> $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

<u>the answer is</u>

$x^{2}-2x-4=0$

5 0

3 years ago

Read 2 more answers

Find the missing dimension of the following parallelogram.

Scrat [10]

Answer:

21 (PLEASE GIVE BRAINLIEST)

Step-by-step explanation:

we have the base, but we need the height

we will find it using the area:

height = 336 / 16 = 21

7 0

3 years ago

(x-3)^2=-12 using square root

Anna [14]

Sqrt((x-3)^2) = sqrt(-12)
x-3=<span>±</span>sqrt(-12)

8 0

3 years ago