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BabaBlast [244]

3 years ago

14

This tennis ball has an diameter of 2.7 inches. What is the volume of this tennis ball? Use 3.14 for π and round your answer to

a whole number.

1 answer:

Pani-rosa [81]3 years ago

8 0

Answer:

V = 10 in^3

Step-by-step explanation:

The volume of a sphere is given by

V = 4/3 pi r^3

We have the diameter

r = d/2 = 2.7/2 =1.35

V = 4/3 ( 3.14) (1.35)^3

V =10.30077

Rounding to the nearest whole number

V = 10 in^3

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3 (x - 3) + x (r + 4)

docker41 [41]

Answer:

Step-by-step explanation:

3(x-3)+x(r+4)

3x-9+xr+4x

7x+9+xr

7 0

3 years ago

Help is very much needed :)

ioda

Answer:

6

Step-by-step explanation:

10/5 ×/3 multiply 10×3 and than divide by 5

or another way start with multiplying 5 and n equaling 5n=

than 10× 3= 30 so you're left with 5n=30 divied 30 by 5 equalling 6

so x=6

3 0

3 years ago

(Solve and write addition Equations) Zacarias and Paz together have $756.80. If Zacarias has $489.50, how much does Paz have? Wr

serious [3.7K]

$489.50+x=$756.80

x=amount paz has.

756.80-489.50 = 267.3

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4 0

3 years ago

Read 2 more answers

What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)

Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = $L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\$

Now to solve the integral on the right hand side we shall use Integration by parts Taking $7t^{3}$ as first function thus we have

$\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\$

Again repeating the same procedure we get

$=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\$

Again repeating the same procedure we get

$\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\$

Now solving this integral we have

$\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}$

Thus we have

$L[7t^{3}]=\frac{7\times 3\times 2}{s^4}$

where s is any complex parameter

5 0

3 years ago

Select from the drop-down menu to correctly identify the property shown.

tatuchka [14]

Commutative Property. I did this last year. Trust me

7 0

3 years ago

Read 2 more answers