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2 years ago

I need this question done thanks

Mathematics

1 answer:

Fudgin [204]2 years ago

6 0

Answer:

H. 40%

Step-by-step explanation:

18/30 = 60%

So 60% of kids have last names with more than 8 letters.

That means 40% of kids have last names with fewer than 8 letters.

(Because the percentages have to add up to 100)

Hope this helps! :)

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9dx10=240 pls hurry

vaieri [72.5K]

Answer:

2.6 or $\frac{8}{3}$

Step-by-step explanation:

9d×10=240

÷10 ÷10

9d =24

÷9 ÷9

2.6 or $\frac{8}{3}$

8 0

2 years ago

50 POINTS & BRAINLIEST!!!! HELP ASAP!!!

Allisa [31]

6(2x - 11) + 15 = 21

First step use distributive property on the left side to remove the parenthesis, to do this multiply 6 by each term inside the parenthesis:

12x - 66 + 15 = 21

Second step, simplify the left side by combining the like terms:

12x - 51 = 21

Third step, add 51 to both sides of the equation:

12x = 72

Last step, divide both sides by 12:

x = 72/12

x = 6

3 0

3 years ago

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

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