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sergiy2304 [10]
3 years ago
12

HELP ASAP!!!!! HELP FAST!!!!!

Mathematics
1 answer:
mr_godi [17]3 years ago
8 0

Answer:

no they are not because 11, 14 is the only one that has a vertices

Step-by-step explanation:

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What is the solution of x+8/5x-1>0?
DerKrebs [107]

Answer:

x<-8 or x>1/5

Step-by-step explanation:

\frac{x+8}{5x-1} >0

First we replace > symbol by = sign

\frac{x+8}{5x-1} =0

To solve for x we multiply 5x-1 on both sides

x+8 =0

x=-8

5x-1=0 solve for x

x= 1/5

WE got two values x=-8  and x= 1/5

Now we make a number line and check with each interval

x<-8,  -8<x<1/5,  x>1/5

Pick x= -9 and check with the given inequality

\frac{-9+8}{5(-9)-1} >0

1/46 >0 is true, so x<-8 satisfies our inequality

Pick x= 0 and check with the given inequality

\frac{0+8}{5(0)-1} >0

-8 >0 is false , -8<x<1/5 does not satifies our inequality

Pick x= 2 and check with the given inequality

\frac{2+8}{5(2)-1} >0

10/9>0 is true , x>1/5  satisfies our inequality

3 0
3 years ago
Read 2 more answers
The area of a square is 729 square feet.What is the length of the side?
Art [367]

To find the length of the side of a square when the area is known, find the square root of the area:


Side length = √729

Side Length = 27 feet.

7 0
4 years ago
Cj took a math test and got 36 correct and 9 incorrect answers. What was percentage of correct answers?
Anastaziya [24]

Answer:

Step-by-step explanation:

Your answer would be 80% or a B

4 0
3 years ago
Find the centroid for an area defined by the equations y = x² + 3 and y = - (x – 2)² + 7
sergeinik [125]
First we need to find where the 2 graphs intercept.

x^2 + 3 = - (x^2 - 4x + 4) + 7
x^2 + 3 = -x^2 + 4x + 3
2x^2 - 4x = 0 
2x(x - 2)

x = 0 , 2. are x coordinates  of the 2 intercepts.

6 0
3 years ago
Find an ordered pair x-2y=4​
Vesnalui [34]
Answer:

Step by Step:
x-2y=4
+2y +2y
x=-2y+4

x-2y=4
-4 -4
x-2y-4=0
+2y +2y
x-4=2y
Divid by 2
1/2x-2= y

x=-2y+4
x=-2(1/2x-2)+4
x=-x+4+4
x=-x+8
+x +x
2x= 8
x=4

y=1/2x-2
y=1/2(4)-2
y=2-2
y=0
(4,0)
5 0
3 years ago
Read 2 more answers
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