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vodomira [7]
2 years ago
15

Gabrielle is 9 years younger than Mikhail. The sum of their ages is 75 . What is Mikhail's age?

Mathematics
1 answer:
Damm [24]2 years ago
6 0

Answer: We know Gabrielle is 9 yrs younger than Mikhail hence it will be x-9 yrs. Sum of their ages is 87. Mikhail's age is 48 yrs.

HOPE THIS HELPS

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Formula for 4,9,14 19 nth term ​
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Answer:

first get their difference and then get positions of the difference and get the factors and then you will get the nth term

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A circular pool has a diameter of 40 feet and is surrounded by a wooden deck that has a width of 4 feet and a depth of 6 inches.
Dvinal [7]
40x4x6= 960 square inches
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3 years ago
What is the value of x
borishaifa [10]
x^2=4^2+4^2\\
x^2=2\cdot16\\
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5 0
3 years ago
Annie is framing a photo with a length of 6 inches and a width of 4 inches. The distance from the edge of the photo to the edge
Ede4ka [16]

Answer:

Part a) The quadratic function is 4x^{2} +20x-39=0

Part b) The value of x is 1.5\ in

Part c) The photo and frame together are 7\ in wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0

Part b) What is the value of x?

Solve the quadratic equation 4x^{2} +20x-39=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem

we have

4x^{2} +20x-39=0

so

a=4\\b=20\\c=-39

substitute in the formula

x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}

x=\frac{-20(+/-)\sqrt{1,024}} {8}

x=\frac{-20(+/-)32} {8}

x=\frac{-20(+)32} {8}=1.5\ in  -----> the solution

x=\frac{-20(-)32} {8}=-6.5\ in

Part c) How wide are the photo and frame together?

(4+2x)=4+2(1.5)=7\ in

5 0
3 years ago
Please help and tell me why u got that answer, I will mark brainliest
gregori [183]
The answer is C

i think so
4 0
3 years ago
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