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3 years ago

Please answer correctly I will mark you Brainliest!

Mathematics

1 answer:

BartSMP [9]3 years ago

7 0

Answer:

1047.12in³

Step-by-step explanation:

The volume of the sphere is:

V= 4/3 πr³

Where r is radius

Therefore the volume of each pinata is:

V= 4/3 x π x 5³

V= 523.6in³

The total:

V = 523.6 + 523.6 = 1047.12in³

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What Is 4x+6y=48 and what is the general form and x intercept and y intercept and slope and factored form?

Vladimir [108]

Given equation : 4x+6y=48 .

The general form of the linear equation is ax +by =c.

The given equation is already in general form.

Let us find x-intercept.

Plugging y =0 in given equation.

4x+6(0)=48 .

4x +0 = 48

4x = 48

Dividing both sides by 4, we get

4x/4 = 48/4

x = 12.

<h3>Therefore, x-intercept is (12,0).</h3>

Let us find y-intercept.

Plugging x =0 in given equation.

4(0)+6y=48 .

6y = 48

Dividing both sides by 6, we get

6y/6 = 48/6

y = 8.

<h3>Therefore, y-intercept is (0,8).</h3>

Let us convert given equation in slope-intercept form y=mx+b.

4x+6y=48

Subtracting 4x from both sides, we get

4x-4x+6y=48-4x

6y = -4x+48

Dividing both sides by 6, we get

6y/6 = -4x/6+48/6

y = -2/3 x +8.

<h3>Therefore, slope is -2/3.</h3>

Let us find factored form now.

GCF of the given linear equation is 2.

Factoring out 2 in all term, we get

3(2x+3y=24)

That is factored form.

4 0

3 years ago

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

3 years ago

Meredith bought a book that cost 18 at a discount of 16% what did she pay for the book

s344n2d4d5 [400]

I think it is $15.12

I hope that helps!!!!!!

: )

5 0

4 years ago

Read 2 more answers

Someone plz help!!!!!

Gnoma [55]

Need more information on it to solve

5 0

3 years ago

Read 2 more answers

Help help please help help

yuradex [85]

Answer:

The range is 5.

Step-by-step explanation:

You can deduce this by concluding all other statements are true, or by calculating the range. 10 - 1 = 9

3 0

3 years ago