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iren [92.7K]
3 years ago
11

AhhhhhHH please help me

Mathematics
1 answer:
lakkis [162]3 years ago
3 0
Use the y value in equation one and substitute it in equation two.

y=4x-3
y=-x-13

4x-3=-x-13
solve for x
add x to both sides
5x-3=-13
add 3 to both sides
5x=-10
divide both sides by 5
x=-2

Use x=-2 to solve for y in either equation
y=4x-3
y=4(-2)-3
y=-8-3
y=-11

(x, y) = (-2, -11)
Top right is the answer

Hope this helps! :)
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Charley and Sarah went to the candy store.
Pepsi [2]

Answer:

rtt6655r55

Step-by-step explanation:

4 0
4 years ago
What are the solutions to the equation below?
Kryger [21]
So, as you can see, there are actually two equations here:
x- 4 = 0 and 8x+64=0
So, we need to solve both of them to get the solution.

x-4=0
x = 4

8x+64=0
8x=-64
x=-64/8
x=-8

So, the correct answer is C. x=4 and x=-8

4 0
3 years ago
NOTE: I FINISHED THE FIRST PART, I JUS NEED HELP W THE SECOND PART BELOW!!!
bagirrra123 [75]

Answer:

  see below

Step-by-step explanation:

<em>Which of the equations from part A represent adding two rational numbers?</em>

  Equations A, C, E

<em>What hypothesis can you make about the sum of two rational numbers?</em>

  The sum of two rationals will always be rational

<em>Will the addition result in a rational or an irrational number?</em>

  Our hypothesis is that the result is always rational. This can be justified by the fact that the sum of two rationals a/b + c/d, where a, b, c, d are integers and bd≠0, is (ad+bc)/(bd), a rational, based on closure of integers for multiplication and addition.

<em>Which equations represent the sum of a rational and an irrational number?</em>

  Equations B, F

<em>What hypothesis can you make about the sum of an irrational and a rational number?</em>

  The sum of a rational and irrational number is always irrational.

4 0
4 years ago
Write a function for the graph described as a transformation of y=x^2.
Pie

we are given

parent function as

y=x^2

(1)

vertical stretch of factor 2

so, we can multiply y-value by 2

y=2x^2

then a shift right of 3 units

so, we can replace x as x-3

y=2(x-3)^2

(2)

a shift left by 2 units

we can replace x as x+2

y=(x+2)^2

then a horizontal shrink factor of 1/2

so, we can multiply by 2 to x-value

y=(2x+2)^2

then a shift down of 5 units

we can subtract y-value by 5

so, we get

y=(2x+2)^2-5

(3)

a shift to the right 1 unit

we can replace x as x-1

y=(x-1)^2

stretched vertically by a factor of 1/2

we can multiply y-value by 1/2

y=\frac{1}{2} (x-1)^2

then is shifted down 4 units

we can subtract y-value by 4

y=\frac{1}{2} (x-1)^2-4

(4)

reflected across the x-axis

we can multiply y-value by -1

y=-x^2

tretched vertically by a factor of 3

multiply y-value by 3

y=-3x^2

shifted left 7 units

we can replace x as x+7

y=-3(x+7)^2


4 0
3 years ago
Solve for x.<br><br> −ax + 2b &gt; 8
Rudik [331]
<span>−ax + 2b > 8
</span>⇒ -ax> 8-2b
⇒ x< (8-2b)/(-a)
<span>
The final answer is </span>x< (8-2b)/(-a)~
7 0
4 years ago
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