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jarptica [38.1K]

3 years ago

13

What two things are needed to create an algebraic equation? Provide an example of an algebraic equation.

1 answer:

iren2701 [21]3 years ago

7 0

Answer:

a variable and an equal sign

Step-by-step explanation:

example: 2x - 3 = 7

2x = 10

x = 5

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A rocket is launched straight up from the ground with an initial velocity of 192 feet per second. The equation for the height of

Ivenika [448]

The equation for the height of the rocket at time t given

$h= -16t^2+192t$

We have to find the time t, when the rocket reaches 560 feet.

That means we have to find t when h = 560 ft. we will place 560 in the place of h to find t now.

$h= -16t^2+192t$

$560 = -16t^2+192t$

In the right side, we can check -16 is the common factor. So we will take out -16 from the rigbht side.

$560 = -16(t^2 - 12t)$

To get rid of -16 from the right side and move it to left side, we will divide both sides by -16.

$560/-16 = -16(t^2-12t)/-16$

$-35 = t^2 -12t$

Now we will move -35 to the righ side by adding 35 to both sides.

$-35+35 = t^2-12t+35$

$0 = t^2 -12t+35$

$t^2-12t+35 = 0$

We will factorize thee left side to find the values of t now. We need to find a pair of factors of 35 that by adding them we will get -12.

The pair of factors of 35 are -5 and -7 and by adding -5-7 we will get -12.

$t^2-12t+35 =0$

$(t-5)(t-7) =0$

So by using zero product property we will get

$t-5 =0$

$t-5+5 = 0+5$

$t=5$

Also $t-7 =0$

$t-7+7 = 0+7$

$t=7$

So we have got the rocket reaches at 560ft when t = 5 seconds and also when t = 7 seconds.

Now part b.

When the rocket completes its trajectory and hits the ground then the height or h = 0. So we will place h = 0 there in the equation.

$h= -16t^2+192t$

$0= -16t^2 + 192 t$

$0 = -16(t^2-12t)$

$-16(t^2-12t) = 0$

We will move -16 to the other side by dividing it to both sides.

$-16(t^2-12t)/-16 = 0/-16$

$t^2-12t = 0$

We will take out the common factor t from the left side. By taking out t we will get,

$t(t-12) = 0$

We will use zero product property now. By using that we will get,

$t = 0$

ans also $t-12 = 0$

$t-12+12 = 0+12$

$t = 12$

When the rocket completes its trajectory and hits the ground the time t can not be 0. When t =0, the rocket starts the trajectory.

So when the rocket completes its trajectory and hits the ground ,

then t = 12seconds.

So we have got the required answers.

6 0

3 years ago

13 Levi bought a new phone for $550. He received a discount of 24%

Likurg_2 [28]

After discount: $418; total (including tax): $388.74

7 0

3 years ago

PLEASE HELP!!!<br> Find the common difference for the arithmetic sequence. -55, -50, -45, -40.....

Fed [463]

Answer:

5

Step-by-step explanation:

Just grap any two consecutive terms in the sequence. lets say -55 and -50

-50-(-55) = 5

you can check this by picking two other terms

7 0

3 years ago

Rewrite the equation in Ax+By=C form

Ksju [112]

Ax + By = C form of the given equation is –6x + y = –28.

Solution:

Given equation is y – 2 = 6(x – 5).

To write the equation in Ax + By = C format:

y – 2 = 6(x – 5)

y – 2 = 6x – 30

Add 2 on both sides of the equation.

y – 2 + 2 = 6x – 30 + 2

y = 6x – 28

Subtract 6x from both sides of the equation.

y – 6x = 6x – 28 – 6x

y – 6x = –28

Arrange the terms in the equation.

–6x + y = –28

This is in the form of Ax + By = C.

Here A = –6, B = 1 and C = –28.

3 0

3 years ago

Find the length of the segment indicated.<br><br> A. 16.4<br> B. 11.4<br> C. 12.1<br> D. 13.3

kykrilka [37]

using Pythagorean triplet

$\\ \sf\longmapsto P^2=H^2-B^2$

$\\ \sf\longmapsto x^2=19.6^2-15.4^2$

$\\ \sf\longmapsto x^2=384.16-237.16$

$\\ \sf\longmapsto x^2=147$

$\\ \sf\longmapsto x=\sqrt{147}$

$\\ \sf\longmapsto x=12.1$

4 0

3 years ago