Point D is on segment BC. Segment BC measures 8x

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 the first step in solving the equation 3.5 n + 6.4 = 42.5?

miss Akunina [59]

I would add like terms, meaning subtract 6.4 from both sides.

3.5n = 36.1

7 0

3 years ago

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2/3x+2 -x-3 <br> solution and the steps how to do it

Stells [14]

$\text{Hey there!}$

$\mathsf{\dfrac{2}{3}x+2-x-3}\\\\\mathsf{COMBINE\ your\ like\ terms}\\\\\mathsf{(\dfrac{2}{3}x-x)+(2-3)}\\\\\mathsf{\dfrac{2}{3}x-x=\dfrac{-1}{3}}\\\\\mathsf{2-3=-1}\\\\\boxed{\boxed{Thus,\ your \ answer\ is: \boxed{\dfrac{-1}{3}x -1}}}\checkmark$

$\text{Good luck on your assignment and enjoy your day!}$

~ $\frak{LoveYourselfFIrst:)}$

7 0

3 years ago

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Heyyyyy pleasse help me please please and thank you

dimulka [17.4K]

Answer:

x ≈ 33.1°

Step-by-step explanation:

Using the tangent ratio in the right triangle

tan x = $\frac{opposite}{adjacent}$ = $\frac{125}{192}$ , then

x = $tan^{-1}$ ( $\frac{125}{192}$ ) ≈ 33.1° ( to the nearest tenth )

7 0

2 years ago

Use the quadratic formula to solve x2 + 9x + 10 = 0.<br> What are the solutions to the equation?

vampirchik [111]

Answer:

$\large\boxed{x=\dfrac{-9\pm\sqrt{41}}{2}}$

Step-by-step explanation:

$\text{The quadratic formula of}$

$ax^2+bx+c=0$

$\text{If}\ b^2-4a0,\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\=========================================$

$\text{We have}\ x^2+9x+10=0\\\\a=1,\ b=9,\ c=10\\\\\text{substitute:}\\\\b^2-4ac=9^2-4(1)(10)=81-40=41>0\qquad _{\text{two solutions}}\\\\\sqrt{b^2-4ac}=\sqrt{41}\\\\x=\dfrac{-9\pm\sqrt{41}}{2(1)}=\dfrac{-9\pm\sqrt{41}}{2}$

5 0

3 years ago

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