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3 years ago

What is the answer to the following equation 13pt13-p

Mathematics

1 answer:

xeze [42]3 years ago

8 0

You have to distribute it.
EXPLANATION:

13 pt 13-p
13-13
0 pt-p
-p-0
-p= 14
14 pt
= 14

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Where would i draw my reflection

Maurinko [17]

Answer:

The coordinates for the reflection of quadrilateral ABCD is..

A'(-7,2) ----> A'(-7,-2)

B'(-7,4) ----> B'(-7,-4)

C'(-4,6) ----> C'(-4,-6)

D'(-1,6) ----> D'(-1,-6)

Step-by-step explanation:

Since it's being reflected over the x-axis, the x remains the same and the y becomes a negative.

3 0

3 years ago

Find the value of 2a² + 5b² when a= -6 and b= 2

sashaice [31]

2(-6^2)+5(2^2)
2(36)+5(4)
72+20
92

4 0

3 years ago

Read 2 more answers

Patel is solving 8x2 + 16x + 3 = 0. Which steps could he use to solve the quadratic equation? Check all that apply. 8(x2 + 2x +

BlackZzzverrR [31]

Answer:

Patel can use the first and last steps.

Step-by-step explanation:

The given equation is :

$8x^{2} +16x+3=0$

Patel can use the last step.

$8(x^{2} +2x)+3=0$

Taking 3 to right hand side, we get

$8(x^{2} +2x)=-3$

He can also use this step.....

$8(x^{2} +2x+1)=-3+8$

Upon solving we get,

$8x^{2} +16x+8=-3+8$

=> $8x^{2} +16x=-3+8-8$

=> $8x^{2} +16x=-3$

7 0

4 years ago

Solve the following system.

Brilliant_brown [7]

<h3>The solutions are:</h3>

$(0, -1)\ and\ (\frac{1}{3}, 0)$

<em><u>Solution:</u></em>

<em><u>Given that,</u></em>

$y = 3x^2 + 2x-1-------- eqn 1\\\\\3x-y=1 ------- eqn 2$

From eqn 2,

3x - y = 1

y = 3x - 1

<em><u>Substitute the above in eqn 1</u></em>

$3x -1 = 3x^2 + 2x - 1\\\\3x^2 + 2x - 3x =0\\\\3x^2 -x = 0\\\\x(3x - 1) = 0\\\\Therefore,\\\\x = 0\\\\And\\\\3x - 1 = 0\\\\3x = 1\\\\x = \frac{1}{3}$

Substitute x = 0 in eqn 2

3(0) - y = 1

y = -1

Substitute x = 1/3 in eqn 2

$3\frac{1}{3} - y = 1\\\\ 1 - y = 1\\\\y = 0$

Thus solutions are:

$(0, -1)\ and\ (\frac{1}{3}, 0)$

4 0

4 years ago

What plus what equals 1,250

mamaluj [8]

625+625=1250
hope this helped

4 0

3 years ago