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elena55 [62]
3 years ago
12

What are the zeros of f(x) = x2 - 8x+16?

Mathematics
1 answer:
Stels [109]3 years ago
3 0

Answer:

x=4

Step-by-step explanation:

f(x) = x^2 - 8x+16

Set equal to zero

0 = x^2 -8x +16

Factor

what 2 numbers multiply to 16 and add to -8

-4*-4 = 16

-4+-4 = -8

0= (x-4)(x-4)

Using the zero product property

x-4 = 0  x-4 =0

x=4  x=4

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Simplify: 36÷(–3) + 2(27÷3 x -2).
jeka94

Answer:

-48

Step-by-step explanation:

36/-3 = -12

-12 + 2(27/3 * -2)

27/3 = 9

-12 + 2(9 * -2)

9 * -2 = -18

-12 + 2(-18)

2 * -18 = -36

-12 + -36 = -12 - 36

-48

4 0
3 years ago
How can the graph of f(x) = 1/(- x) - 9 be obtained from the graph of y = 1/x
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2 years ago
The circumference of a circle is 20π cm. What is the DIAMETER of the circle?
Sladkaya [172]

Answer:20cm

Step-by-step explanation:

Circumference=20π

Diameter=circumference ➗ π

Diameter=20π ➗ π

Diameter=20

Diameter=20cm

7 0
3 years ago
What is the correlation coefficient for the data shown in the
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Answer:

1

Step-by-step explanation:

7 0
3 years ago
Graph the functions on the same coordinate axis. {f(x)=−2x+1g(x)=x2−2x−3
NikAS [45]

Answer:

(2,-3) and (-2,5)

Step-by-step explanation:

Let us graph the two equations one by one.

1. f(x)=-2x+1

If we compare this equation with the slope intercept form of a line which is given as

y=mx+c

we see that m = -1 and c =1

Hence the slope of the line is -2 and the y intercept is 1. Hence one point through which it is passing is (0,1) .

Let us find another point by putting x = 1 and solving it for y

y=-2(1)+1

y=-2+1 = -1

Let us find another point by putting x = 2 and solving it for y

y=-2(2)+1

y=-4+1 = -3

Hence the another point will be (2,-3)

Let us find another point by putting x = -2 and solving it for y

y=-2(-2)+1

y=+1 = 5

Hence the another point will be (-2,5)

Now we have two points (0,1) ,(1,-1) ,  (2,-3) and (-2,5) we joint them on line to obtain our line  

2.

g(x)=y=x^2-2x-3

y=x^2-2x+1-1-3

y=(x-1)^2-4

(y+4)=(x-1)^2

It represents the parabola opening upward with vertices (1,-4)

Let us mark few coordinates so that we may graph the parabola.

i) x=0 ; y=y=(0)^2-2(0)-3=0-0-3=-3 ; (0,-3)

ii)x=-1 ; y=(-1)^2-2(-1)-3=1+2-3=0 ; (-1,0)

iii) x=2 ; y=(2)^2-2(2)-3 = 4-4-3 =-3 ;(2,-3)

iii) x=1 ; y=(1)^2-2(1)-3 = 1-2-3 =-4  ;(1,-4)

iii) x=-2 ; y=(-2)^2-2(-2)-3 = 4+4-3 =5  ;(-2,5)

Now we plot them on coordinate axis and line them to form our parabola

When we plot them we see that we have two coordinates (2,-3) and (-2,5) are common , on which our graphs are intersecting. These coordinates are solution to the two graphs.

3 0
3 years ago
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