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3 years ago

Solve the equation y + 5=-12 A y=-19 B y=-17 C y=-7 D y=7

Mathematics

2 answers:

mrs_skeptik [129]3 years ago

8 0

Answer:

y=17

Hope this helps you

zalisa [80]3 years ago

6 0

B. Y= -17
Since it’s a negative plus a positive you’re pretty much subtracting.

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Kiran says that a solution to the equation x+4=20 must also be a solution to the equation 5(x+4)=100 Write a convincing explanat

Helen [10]

Answer:

x+4=20 is a solution to 5(x+4)=100.

Step-by-step explanation:

Following PEMDAS, divide 5 from both sides of the equation 5(x+4)=100, and you have x+4=20 remaining.

This means that both equations have the same solution.

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3 years ago

Which undefined terms are needed to define a line segment

Finger [1]

Unsure of what you are asking!

But if the issue here is how to define a line segment, write what you do know and then reconsider "undefined terms."

A line segment is a straight line that connects a given starting point and given ending point.

If you consider a circle of radius 3 units, the radius can be thought of as the line segment connecting the center of the circle to any point on the circumference of the circle.

If the center of a given circle is at C(0,0) and a point on the circumference is given by R(3sqrt(2),3sqrt(2)), then AC is the line segment joining these two points. This line segment has length 3 and is in the first quadrant, with coordinates x=3sqrt(2) and y=3sqrt(2) describing the end point of the segment.

3 0

3 years ago

Read 2 more answers

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
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7 0

4 years ago

Can someone help me on this

masha68 [24]

The answer to this question is 120winners

6 0

3 years ago

Helppppppppppppppppppppppppppppp

evablogger [386]

Answer: 33

Step-by-step explanation:

(f of g)(2) = f(g(2))

g(2) = -2(2) = -4

f(-4) = (-4)^2 - 3(-4) + 5 =

16 - - 12 + 5 =

16 + 17 =

7 0

1 year ago