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iogann1982 [59]
3 years ago
14

The endpoints of one diameter of a circle are (9,3) and (-5,15). What is the center of the circle?

Mathematics
1 answer:
fiasKO [112]3 years ago
7 0

Answer:

(2,9)

Step-by-step explanation:

X midpoint:9+(-5)/2 which gives 2

Y midpoint:3+15/2 which gives 9

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Laura delivers newspapers. She spent 4|12 of her savings one new CD. Which equivalent fraction shows the amount Laura spent?
julia-pushkina [17]
Equivalent fractions are fractions that have the same value. They can be found by reducing the fraction or by multiplying both the numerator and denominator by the same number.

In this case, 4 and 12 can both be divided by 4, so the fraction 4/12 can be reduced to 1/3.
6 0
3 years ago
Please help me to prove this!​
Sophie [7]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B = C                → A = C - B

                                          → B = C - A

Use the Double Angle Identity:     cos 2A = 2 cos² A - 1

                                             → (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

<u>Proof LHS → RHS:</u>

LHS:                     cos² A + cos² B + cos² C

\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C

\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C

\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C

\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]

\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)

\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)

\text{Even/Odd:}\qquad \qquad 1+2\cos C \cdot \cos A\cdot \cos B\\\\\\.\qquad \qquad \qquad \quad =1+2\cos A \cdot \cos B\cdot \cos C

LHS = RHS: 1 + 2 cos A · cos B · cos C = 1 + 2 cos A · cos B · cos C   \checkmark

5 0
3 years ago
Simplify each expression as much as possible, and rationalize denominators when applicable. √72=?
Misha Larkins [42]

Answer:

6√2

Step-by-step explanation:

Solving the given expression step by step:

72 = 2 × 2 × 2 × 3 × 3

Now for finding the square root we will make pair of two numbers of same values.

We have one pair of 2's and one pair of 3's and left one 2

Thus, √72 = 2 × 3 × √2 = 6√2

We rationalize denominator and change it into a simpler form as soon as possible.

3 0
3 years ago
Graph the line with slope 2/3 passing through the point (-1,-2).
Vsevolod [243]

Answer:

1/2

Step-by-step explanation:

u will need to use desmos but i beleive u evaluate

3 0
3 years ago
Please answer whats in the screenshot ASAP pleasee thank you so so much have a great day! &lt;333 :)
maria [59]

Answer:

3

Step-by-step explanation:

I believe it is 3 because you are multiplying each number by 3 to get your bigger shape

5 0
3 years ago
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