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Natali5045456 [20]
3 years ago
6

I needd the whole answer for it​

Mathematics
2 answers:
natita [175]3 years ago
5 0
What is the question we need one?
Radda [10]3 years ago
3 0
Of what? You didn’t write the question honey
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Since a math school opened the girls basketball team had the same record every season the team won a total of 182 games while lo
Tems11 [23]

Answer:

11

Step-by-step explanation:

I guess

6 0
3 years ago
274,635 rounded to the nearest thousands
Firlakuza [10]

275000 is your answer

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3 years ago
Guide
Tema [17]

Answer:

Step-by-step explanation:

4 0
3 years ago
Please help!
agasfer [191]
Ok im not sure the answer yet so ima work on it at the same time while im explaining it. (The answer will probably be near the end)

We can use the elimination method to eliminate y out. To do that we multiply the first equation by 3.

6x+3y=-12

Now just subtract it from the other equation.

6x+3y=-12
5x+3y=-6

***x=-6***

Usually after doing the elimination method you will have to solve for x but in this case its already solved for you. If you want to find y now you just take the first equation and fill x with -6 and solve for y.

2(-6)+y=-4
-12+y=-4
y=8

Brainliest my answer if it helps you out?
4 0
4 years ago
Read 2 more answers
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
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