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Svetach [21]
3 years ago
7

Which graph does NOT represent a function?

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
3 0

Answer:

The circle graph does not represent a function because it doesn't pass the vertical line test.

Step-by-step explanation:

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I NEED HELP FAST PLEASE! Find the surface area of the following composite figure.
jekas [21]

the answer is 228 I was wrong when I said

I think the choices is wrong

3 0
3 years ago
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A number line contains points Q, R, S, and T. Point Q is on the coordinate 24, R is on the coordinate 28, S is on the coordinate
Alex777 [14]

Answer:

  72%

Step-by-step explanation:

QT has length 42-24 = 18.

ST has length 42-29 = 13.

The length ST is 13/18 ≈ 72.2% of the length of QT.

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3 years ago
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Ava wants to draw a parallelogram on the coordinate plane. She
wolverine [178]

Answer:

k =(2,1)

JK = 2

Step-by-step explanation:

Given

J = (0,1) ---- (x_1,y_1)

H = (1,-2) --- (x_2,y_2)

I = (3,-2) --- (x_3,y_3)

See attachment for grid

Solving (a): The coordinates of K

The parallelogram has the following diagonals: IJ and HK

Diagonals bisect one another. So:

Midpoint of IJ = Midpoint of HK

This gives:

\frac{1}{2}(I + J) = \frac{1}{2}(H+K)

\frac{1}{2}(x_3+x_1,y_3+y_1) = \frac{1}{2}(x_2+x,y_2+y)

\frac{1}{2}(3+0,-2+1) = \frac{1}{2}(1+x,-2+y)

\frac{1}{2}(3,-1) = \frac{1}{2}(1+x,-2+y)

Multiply through by 2

(3,-1) = (1+x,-2+y)

By comparison:

1 + x = 3

-2 + y = -1

Solve for x and y

x = 3 - 1 =2

y = -1 +2 = 1

So, the coordinates of k is:

k =(2,1)

The length of JK is calculated using distance (d) formula

d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2

J = (0,1) ---- (x_1,y_1)

k =(2,1) ---- (x_2,y_2)

So:

d = \sqrt{(0 - 2)^2 + (1 - 1)^2

d = \sqrt{(- 2)^2 + (0)^2

d = \sqrt{4 + 0

d = \sqrt{4

d = 2

Hence:

JK = 2

8 0
3 years ago
Can you guys do this ASAP, thanks will mark brainliest
dybincka [34]

Answer:

A) Graph an quadratic equation for a parabola (x being to the power of 2), and a linear equation. Examples would be y=x² and y=x.

B) Parabolas that share the same vertex can simply have another transformation done to them. For example: y= 5x² and y=3x².

C) A line that would pass through these would be any linear graph WITHOUT a 'b' value, so a y intercept of 0. An example would be y=5x.

5 0
3 years ago
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F(x) = x4 - 32x2 + 7
xenn [34]
F is increasing if f '(x) = 4x^3<span> - 64x >0, so it is the same of  x(x^2 -16)>0, implies x>0 or (x-4)(x+4)>0, so x> + or -4
the answer is </span><span>(-4 , 0) ∪ (4 , infinity)

</span>f is decreasing if f '(x) = 4x^3 - 64x <0, so it is the same of  x(x^2 -16)<0, implies x<0 or (x-4)(x+4)<0, so x< + or -4
the answer is (- infinity, -4) ∪ (0 , 4)

<span>the local minimum and maximum
</span>f '(x) =0, impolies x=+ or -4, or x=0
or f'(o)=0,  and  f'(- 4)=f'(4)= 0, 
M(-4, 0) or M(4, 0) or M(0,0)

<span>inflection points can be found by solving f '' (x)=12x^2 - 64 =0
</span>x=+ or - 4sqrt(3) / 3
so the inflection point is  and M(- 4sqrt(3) / 3, f'' ( -4sqrt(3))  (smaller x value), and M(4sqrt(3) / 3, f'' (4sqrt(3)) (larger x value)

f is concave up if f ''>0
it means 12x^2 - 64>0, so the interval is (- infinity, -4sqrt(3) U (4sqrt(3), infinity)

f is concave down if f ''<0
it means 12x^2 - 64<0 so the interval is (-4sqrt(3)) U (4sqrt(3))

4 0
3 years ago
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