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3 years ago

What is the volume of a sphere with a diameter of 10 m, rounded to the nearest tenth of a cubic meter

Mathematics

1 answer:

kirill [66]3 years ago

4 0

Answer:

4188.8 cm^3

Step-by-step explanation:

We can calculate the volume of a sphere by using the following equation:

$\frac{4}{3}\pi r^{3}$

(r=radius)

so let's just plug in what we know

4/3(pi)(10)^3

we can calculate this with a calculator to get 4188.8 cm^3.

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Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$

Since time can only be positive we reject the $t_{2}$ solution and we keep that Ben's book took $t=4.3276$ seconds to reach the ground.

Similarly solving for Josh we obtain

$t_{1}=3.6794sec\\t_{2}=-0.6794sec$

Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

So to answer the original question:

Whose textbook reaches the ground first and by how many seconds?

Josh's textbook reached the ground first
Josh's textbook reached the ground first by a difference of $t=0.6482$

3 0

3 years ago

1. f(6) if f(x) = 4x I know the answer I just need the explanation please thank you!

suter [353]

So when you have one of these, it’s just asking you so solve it as if x were 6 (as you see where the x in f(x) is replaced with a 6.

so you’d sub x for 6

giving you:

4 x (6)

which = 24

hope that makes sense :)

6 0

2 years ago

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Elija, Emily, Edward, and Esme each drew a conclusion about the figure below.

Anvisha [2.4K]

I think the answer is D all of the above

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3 years ago

Which polynomial is prime? x3 + 3x2 – 2x – 6 x3 – 2x2 + 3x – 6 4x4 + 4x3 – 2x – 2 2x4 + x3 – x + 2

alexandr1967 [171]

D. 2x4+x3–x+2 is prime

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