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3 years ago

(-10x^2+9)+(8x^2-5x+7) (−10x 2 +9)+(8x 2 −5x+7)

Mathematics

1 answer:

zavuch27 [327]3 years ago

7 0

Answer:

-2x^2-5x+16

Step-by-step explanation:

-10x^2-8x^2-5x+9+7=-2x^2-5x+16

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Xy+4y=3x-2, show that y is a function of x

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(xy+4y)/x = (3x-2)/x
(y+4y)/x = (3x-2)/x divide both side by x
5y /x = (3x-2)/x
5y= 3x-2 Mult by x both side
y= (3x-2)/5

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3 years ago

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3 years ago

Write 2/3 as a percent. Round your answer to the nearest tenth of a percent.

sweet [91]

divide 2 b 3

2 divided by 3 = 0.625

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2 years ago

The perimeter of the kitchen is 528 inches if the width of the kitchen is 120 inches what is the length of the kitchen

butalik [34]

Answer: 144 inches.

Step-by-step explanation:

1. To solve this problem you must remember the formula for calculate the perimeter of a rectangle, which is shown below:

$P=2l+2w$

Where $l$ is the lenght and $w$ is the width.

2. You know the perimeter and the width, then you can solve for the length, as following:

$P-2w=2l\\l=\frac{P-2w}{2}\\l=\frac{(528in)-(2*120in))}{2}\\l=144in$

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2 years ago

The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

8 0

3 years ago