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Jobisdone [24]
3 years ago
10

If 7x-2y=z then -14x+4y+3z

Mathematics
1 answer:
creativ13 [48]3 years ago
8 0
I'm confused what are you suppose to do
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How many ways can a teacher pick a group of 3 students from his class of 32?
tamaranim1 [39]

Answer:

B

Step-by-step explanation:

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3 years ago
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5.) Juan answered 24/25 correctly on his quiz.<br>What percent of the questions did he get correct?​
NemiM [27]

Answer:

96%

Step-by-step explanation:

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3 years ago
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Rectangle A has a length of 2x + 6 and a width of 3x. Rectangle B has a length of x + 2 and an area of 12 square units greater t
Maurinko [17]
So here is how you solve for the answer.
Firstly, you solve for the Area of Rectangle A.
The formula for Area is Length x width.
So A = (2x + 6)(3x) and the result is: 6x^2 + 18x
Now, let y be the width of rectangle B.
<span>(x+2) (y) = 6x^2 + 18x + 12
(x+2) y = 6(x+1)(x+2)
y = 6(x+1)
</span>So the final answer would be width is 6x + 6. The answer is the third option. Hope this answer helps.
7 0
3 years ago
You have 80 pieces of fabric to make a wall hanging. Each piece of fabric is a square with a side length of 5 inches. You use as
ahrayia [7]
The are would be 25

and we could make 3 squares. 75

5inches will be left over.
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3 years ago
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Explain how to write a quadratic equation given the following three points on the graph (5,31) (3,11) (0,11)
masha68 [24]

Given:

The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).

To find:

The equation of the quadratic function.

Solution:

A quadratic function is defined as:

y=ax^2+bx+c            ...(i)

It is passes through the point (0,11). So, substitute x=0,y=11 in (i).

11=a(0)^2+b(0)+c

11=c

Putting c=11 in (i), we get

y=ax^2+bx+11               ...(ii)

The quadratic function passes through the point (5,31). So, substitute x=5,y=31 in (ii).

31=a(5)^2+b(5)+11

31-11=a(25)+5b

20=25a+5b

Divide both sides by 5.

4=5a+b                  ...(iii)

The quadratic function passes through the point (3,11). So, substitute x=3,y=11 in (ii).

11=a(3)^2+b(3)+11

11-11=a(9)+3b

0=9a+3b

Divide both sides by 3.

0=3a+b                 ...(iv)

Subtracting (iv) from (iii), we get

4-0=5a+b-3a-b

4=2a

\dfrac{4}{2}=a

2=a

Putting a=2 in (iv), we get

0=3(2)+b

0=6+b

-6=b

Putting a=2,b=-6 in (ii), we get

y=(2)x^2+(-6)x+11

y=2x^2-6x+11

Therefore, the required quadratic equation is y=2x^2-6x+11.

7 0
2 years ago
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