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1 year ago

Can u help me with this I don’t understand

Mathematics

1 answer:

Snowcat [4.5K]1 year ago

5 0

Answer:

D -6, 15

Step-by-step explanation:

Because 3,11 is the middle, and 12,7 is the other end, you can do

12 - 3 = 9

3 - 9 = 6

then

11 - 7 = 4

11 + 4 = 15

6 is you x and 15 is your y

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What is the solution to the system of equations? y = 2x + 4 y = -x + 1

den301095 [7]

Step-by-step explanation:

start by setting each equation to each other

$2x + 4 = - x + 1$

first we are going to subtract 2x from both side of the equation

$4 = - 3x + 1$

now subtract 1 from both sides

$3 = - 3x$

now divide by -3 on both sides

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3 years ago

Approximently how many centiliters are in 3 quarts? Round your answer to the nearest unit.

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There should be 283.9. (rounded to the nearest unit). :)

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2 years ago

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In the coordinate plane, what is the length of the line segment that connects points at (5, 4) and (−3, −1) ? Enter your answer

stiks02 [169]

The distance between two points knowing theirs coordinates:

AB =√[(x₂-x₁)² +(y₂-y₁)²]; ===>A(5,-4) & B(-3,-1) Given
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2 years ago

Answers fast please. 1.) y=x−63x+2y=8 Use the substitution method. A.) (4, −2) B.) (14, 8) C.) (0, −6) D.) (3, −3) 2.) What is t

GrogVix [38]

QUESTION 1

The given system of equation is

$y=x-6...eqn1$

and

$3x+2y=8...eqn2$

Let us substitute equation (1) into equation (2) to get,

$3x+2(x-6)=8$

We expand the bracket to get,

$3x+2x-12=8$

We simplify to get.

$5x-12=8$

We group like terms to get

$5x=8+12$

$\Rightarrow 5x=20$

$\Rightarrow x=4$

We now substitute $x=4$ in to equation (1) to obtain,

$y=4-6=-2$

The correct answer is option A.

QUESTION 2

The given system of equations is

$y-x=9...eqn1$

and

$10+2x=-2y...eqn2$

We make y the subject in equation (2) to get,

$y=-x-5..eqn3$

We put equation (3) into equation (1) to obtain,

$-x-5-x=9$

We group like terms to get,

$-x-x=9+5$

This implies that,

$-2x=14$

We divide through by $-2$ to get,

$x=-7$

Hence the x-coordinate is $-7$

QUESTION 3

The given system is

$y=3x-1..eqn1$

and

$x-y=-9...eqn2$

We make $x$ the subject in equation (2) to get,

$x=y-9...eqn3$

We put equation (3) into equation (1) to obtain,

$y=3(y-9)-1$

We expand the bracket to get,

$y=3y-27-1$

Group like terms to get,

$y-3y=-27-1$

We simplify to get;

$-2y=-28$

This implies that,

$y=14$

Therefore the y-coordinate is 14.

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3 years ago

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