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kkurt [141]
2 years ago
14

Circle T is defined by the equation (x + 5)2 + (y - 2) = 81. Which of the following shows the location of the center of Circle T

and the radius of this
circle?
A T is located on (-5, 2); radius = 9 units
B T is located at (5,-2); radius = 9 units
C T is located at (-5, 2); radius = 81 units
D T is located at (5,-2); radius = 81 units
Mathematics
1 answer:
mina [271]2 years ago
4 0

Answer:

A) T is located on (-5, 2); radius = 9 units

Step-by-step explanation:

eqn of circle= (x-a)^2 + (y-b)^2 = r^2

where (a,b) is centre of circle.

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

(y-2)^2 = (y - (2) )^2

(-5,2)

r^2 =81

r = sqrt of 81

r = 9

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Which two numbers on the number line have an absolute value of 2.75?
AysviL [449]

Answer:

2.75 and -2.75

Step-by-step explanation:

The absolute value of 2.75 is 2.75 and the absolute value of -2.75 is 2.75

3 0
2 years ago
Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)
strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

y=a(x-h)^{2} +k

where

(h,k) is the vertex

if a>0 -------> the parabola open upward (vertex is a minimum)

if a -------> the parabola open downward (vertex is a maximun)

<u>case A)</u> f(x)=(x-2)^{2}

This is a vertical parabola open upward

the vertex is the point (2,0)

therefore

The function f(x)=(x-2)^{2}  does not have a vertex on the y-axis

<u>case B)</u> f(x)=x(x+2)

f(x)=x(x+2)=x^{2}+2x

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

f(x)+1=x^{2}+2x+1

Rewrite as perfect squares

f(x)+1=(x+1)^{2}

f(x)=(x+1)^{2}-1

the vertex is the point (-1,-1)

therefore

The function f(x)=x(x+2) does not have a vertex on the y-axis

<u>case C)</u> f(x)=(x-2)(x+2)

f(x)=(x-2)(x+2)=x^{2}-2^{2}

f(x)=x^{2}-4

the vertex is the point (0,-4)

The x-coordinate of the vertex is equal to zero

therefore

The function f(x)=(x-2)(x+2) has a vertex on the y-axis

<u>case D)</u> f(x)=(x+1)(x-2)

f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

f(x)+2= x^{2} -x

Complete the square. Remember to balance the equation by adding the same constants to each side.

f(x)+2+0.25= x^{2} -x+0.25

f(x)+2.25= x^{2} -x+0.25

Rewrite as perfect squares

f(x)+2.25= (x-0.50)^{2}

f(x)=(x-0.50)^{2}-2.25

the vertex is the point (0.5,-2.25)

therefore

The function f(x)=(x+1)(x-2) does not have a vertex on the y-axis

<u>the answer is</u>

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

6 0
3 years ago
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What is the expression for "the difference between 5 times a number and twice that number"?
ycow [4]
Hey there :)

Before translating the sentence into mathematics, let r represent 'a number'

5 times a number = 5r
2 times that number = 2r

The difference between them = 5r - 2r 

* Your equation 
5r - 2r

* Solving the equation 
= 5r - 2r
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Find the difference -11 - -5, 25 - (-3) ,-7 - (-2), 4 - 15, 6 - (-9)
Leviafan [203]

Answer:

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25-(-3)= 28

-7-(-2)= -5

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6-(-9)= 15

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lys-0071 [83]
That would be 5, all you need to do is divide by 2 or mltiply by 0.5 to get half of a number
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