Hi!!!! could someone please help im the middle of taking a test

Solve using substitution-2x-y=-9. 5x-2y=18

Sunny_sXe [5.5K]

Equation 1) -2x-y=-9
Equation 2) 5x-2y=18

Multiply all of equation 1 by 2, so that both equations have (2y) in them.

1) 2(-2x-y=-9)

Simplify.

1) -4x-2y=-18
2) 5x-2y=18

Subtract equations with each other.

-9x=-36

Divide both sides by -9.

x=4

Plug in 4 for (x) into equation 1.

-2x-y=-9

-2(4)-y=-9

Simplify.

-8-y=-9

Add 8 to both sides.

-y=-1

Divide both sides by -1.

y=1

(4, 1)

~Hope I helped!~

5 0

3 years ago

Use the distributive property to expand the expression. −(3x2 + 4y) + (8z2 − 2) A) −3x2 + 4y + 8z2 − 2 B) −3x2 − 4y − 8z2 + 2 C)

otez555 [7]

C) -3x2-4y+8z2-2 is the answer

8 0

3 years ago

In order to form a negation, just add "not" to a statement or take "not" out of a statement if it already exists.

Mnenie [13.5K]

Answer:

The correct answer option for this is: False.

Step-by-step explanation:

We are given the following statement which we are to determine whether its true or false:

In order to form a negation, just add "not" to a statement or take "not" out of a statement if it already exists.

To form a negation, we add the word 'not' to a statement - this part of the given sentence is true.

While 'or take "not" out of a statement if it already exists' is false which makes the whole sentence false.

3 0

4 years ago

Which properties of equality are used to solve the following (in no particular order)?

guajiro [1.7K]

All except combine like terms. Since you only have 1 variable.

Hope this helps.

r3t40

3 0

3 years ago

Plzzz I need help with this question I tried to solve it many times but I can't

blsea [12.9K]

Answer and Step-by-step explanation: Area of a right triangle, (as any other triangle), is calculated as: $A1=\frac{(base)(height)}{2}$

Area of a rectangle is calculated as: $A2=(side)(side)$

Area of a right trapezoid is: $A3=\frac{(a+b)h}{2}$ , where:

a is short base

b is long base

h is height

1) Expressing areas in terms of x:

Area of triangle S1:

$S1=\frac{(2x-3)(4x-6)}{2}$

$S1=4x^{2}-12x+9$

Area of rectangle S2:

$S2 = (4x-6)(3x-2)$

$S2=12x^{2}-26x+12$

Area of trapezoid S3:

$S3=\frac{(2x+3+4x+1)(2x-3)}{2}$

$S3=\frac{(6x+4)(2x-3)}{2}$

$S3=6x^{2}-5x-6$

2) a) $S=4x^{2}-12x+9+12x^{2}-26x+12-(6x^{2}-5x-6)$

$S=4x^{2}-12x+9+12x^{2}-26x+12-6x^{2}+5x+6$

$S=10x^{2}-33x+37$

Which is the same as S = (2x-3)(5x-9)

b) For the areas to be the same:

$\frac{(3x-2+3x-2+2x-3)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}$

$\frac{(8x-7)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}$

$32x^{2}-48x-28x+42=12x^{2}+8x-18x-12$

$20x^{2}-66x+54=0$

Using Bhaskara to solve the second degree equation:

$\frac{66+\sqrt{(-66)^{2}-(4.20.54)} }{2(20)}$

$x_{1}=\frac{66+6}{40}$ = 1.8

$x_{2}=\frac{66-6}{40}$ = 1.5

For the areas of AFGC and ADEB to be equal, x has to be 1.5 or 1.8.

c) Expand a polynomial (or equation) is to multiply all the terms, remiving the parenthesis. Reduce a polynomial (or equation) is to combine terms alike,e.g.:

$S=(2x-3)(5x-9)$