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3 years ago

(PLEASE HELP/LAST QUESTION)

Mathematics

1 answer:

Mashcka [7]3 years ago

8 0

Answer:

Parallel: y=6x-19

Perpendicular: y=-1/6x-0.5

Step-by-step explanation:

To be parallel to a line, the slope must be the same as the line. Then, you plug 3,-1 into y=6x, and add or subtract to balance the equation. You do the same to find the perpendicular line, but the slope must be the negative reciprocal of the line it is perpendicular to.

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Solve the Questions following 81, 8-91, 4pda - Зроби 1. Xut 43 ,

posledela

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4 0

3 years ago

is (3,-3) a solution to the system 5x+4y=3, 2x-5y=19 need to be taught step by step please and thank you.

8090 [49]

Basically all you’re doing is plugging in the inputs to the equations and see if they’re correct.

5(3)+4(-3)=3 this proves to be true
2(3)-5(-3)=21 not 19

8 0

3 years ago

Read 2 more answers

The length of a rectangle is twice the width. The perimeter is 109.8 cm. Find the width and length.

GrogVix [38]

w(2)+w=109.8

2w+w=109.8

3w=109.8

w=36.6

W represents width.

7 0

3 years ago

1. What is the solution to the system of equations?

boyakko [2]

Answer:

$\large\boxed{x=-8\ and\ y=9\to(-8,\ 9)}$

Step-by-step explanation:

$a)\ \text{Elimination method:}\\\\\left\{\begin{array}{ccc}3x+4y=12\\x+2y=10&\text{multiply both sides by (-2)}\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}3x+4y=12\\-2x-4y=-20\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad \boxed{x=-8}\\\\\text{Put the value of x to the second equation:}\\-8+2y=10\qquad\text{add 8 to both sides}\\2y=18\qquad\text{divide both sides by 2}\\\boxed{y=9}$

$b)\ \text{Substitution method:}\\\\\left\{\begin{array}{ccc}3x+4y=12\\x+2y=10&\text{subtract 2y from both sides}\end{array}\right\\\\\left\{\begin{array}{ccc}3x+4y=12&(1)\\x=10-2y&(2)\end{array}\right\qquad\text{subtract (2) to (1)}\\\\3(10-2y)+4y=12\qquad\text{use distributive property}\\(3)(10)+(3)(-2y)+4y=12\\30-6y+4y=12\qquad\text{subtract 30 from both sides}\\-2y=-18\qquad\text{divide both sides by (-2)}\\\boxed{y=9}\\\\\text{Put the value of y to (2):}\\x=10-2(9)\\x=10-18\\\boxed{x=-8}$

5 0

3 years ago

Write in vertex form

nasty-shy [4]

Vertex form: $y=a(x-h)^2+k$

So firstly, put x^2 - 4x into parentheses: $y=(x^2-4x)+6$

Next, we need a perfect square inside the parentheses. Divide the x coefficent by 2 then square that result. In this case, -4 ÷ 2 = -2, (-2)²= 4. Add 4 inside the parentheses and then subtract 4 outside the parentheses:

$y=(x^2-4x+4)+6-4\\y=(x^2-4x+4)+2$

Lastly, factor what's inside the parentheses and your vertex form is $y=(x-2)^2+2$ 

5 0

3 years ago