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3 years ago

10

There is a bag with only red marbles and blue marbles.

1 answer:

zmey [24]3 years ago

4 0

I just need point I apologize

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Which statement is true about the two triangles?

mote1985 [20]

C is the correct answer

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3 years ago

Solve for x 2\3 = 18/x+5

Levart [38]

If you could write out the whole problem exactly how it is, I can solve it for you.

7 0

3 years ago

May anyone please help me?

kiruha [24]

The answer will be D) 8

3 0

3 years ago

Read 2 more answers

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

What is a reasonable estimate for the solution? O (1,-3/4) O (-3/4,1) O (-1,3/4) O (3/4,-1)

Zina [86]

Answer:

See Explanation

Step-by-step explanation:

Your question is incomplete, as the equations or graph or table(s) were not given.

However, I'll give a general way of solving this.

Take for instance, the equations are:

$y = \frac{4}{3}x - 1$

$y = \frac{2}{3}x - \frac{1}{2}$

To do this, we start by equating both equations.

$y = y$

i.e.

$\frac{4}{3}x - 1= \frac{2}{3}x - \frac{1}{2}$

Collect Like Terms

$\frac{4}{3}x - \frac{2}{3}x= 1 - \frac{1}{2}$

Take LCM

$\frac{4x- 2x}{3}= \frac{2 - 1}{2}$

$\frac{2x}{3}= \frac{1}{2}$

Cross Multiply

$2x * 2 = 3 * 1$

$4x = 3$

Make x the subject

$x = \frac{3}{4}$

Substitute 3/4 for x in $y = \frac{4}{3}x - 1$

$y = \frac{4}{3} * \frac{3}{4} - 1$

$y = 1 - 1$

$y = 0$

Hence:

$(x,y) = (\frac{3}{4},0)$

6 0

3 years ago