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3 years ago

What's the answer to this problem

Mathematics

1 answer:

Artyom0805 [142]3 years ago

4 0

So you have to subtract 115 from 15. Then you divide that number by 5 for each relative. That should give you your answer....$20

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What is x?<br> HELP!!<br> Pls

kow [346]

Since line c and d are parallel, line a can act as a transversal. Thus, x=124 degrees because of the alternate exterior angle relationship that can be seen.

7 0

3 years ago

Read 2 more answers

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

X²+y²-6x+14y-1=0<br> and please show your work so I can learn

Eva8 [605]

Hello there,

I hope you and your family are staying safe and healthy during this winter season.

$x^2 + y^2 -6x+14y-1=0$

We need to use the Quadratic Formula*

$x =\frac{-b+\sqrt{b^2}-4ac }{2a}$ , $\frac{-b-\sqrt{b^2} -4ac }{2a}$

Thus, given the problem:

$a = 1, b=-6, c=y^2+14y-1$

So now we just need to plug them in the Quadratic Formula*

$x=\frac{6+2\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$ , $x=\frac{6-\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$

As you can see, it is a mess right now. Therefore, we need to simplify it

$x=\frac{6+2\sqrt{10-y^2-14y} }{2}$ , $x = \frac{6-2\sqrt{10-y^2-14y} }{2}$

Now that's get us to the final solution:

$x=3+\sqrt{10-y^2-14y}$ , $x=3-\sqrt{10-y^2-14y}$

It is my pleasure to help students like you! If you have additional questions, please let me know.

Take care!

~Garebear

3 0

3 years ago

How do I find X in this problem

pogonyaev

You know that BC is congruent to x so you need to solve for x using the ratio:
$\frac{12}{24}=\frac{x}{BC}$

So then we need to find BC.

We know:
$A=\frac{base \times height}{2}$
$28=\frac{BC \times 7}{2}$
Therefore BC =8

Then:
$\frac{12}{24}=\frac{x}{8}$
$x=8 \times \frac{12}{24} = 4$

4 0

4 years ago

Help pleaseeeeee!! D:

Dimas [21]

Answer:

Perimeter = 82 units

Step-by-step explanation:

Find:

Perimeter

Computation:

Unknown height = 13 + x

So,

25² = x² + 24²

x = 7

So,

Unknown height = 13 + 7

Unknown height = 20 unit

Perimeter = 25 + 24 + 13 + 20

Perimeter = 82 units

8 0

3 years ago