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2 years ago

Hw help ASAP PLZZZZZZ

Mathematics

1 answer:

Pachacha [2.7K]2 years ago

3 0

Answer:

Your answer is C. X = 29/8c

Step-by-step explanation:

2/3(cx + 1/2) - 1/4 = 5/2

2cx/3+1/3-1/4=5/2

2cx3+1/12=5/2

2cx/3=5/2-1/12

2cx/3=29/12

(3)2cx/3=29/12(3)

2cx= 31/4

(2c)2cx=29/4(2c)

X=29/8c

Your answer is C. X = 29/8c

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valentina_108 [34]

Answer:

-6

Step-by-step explanation:

-1+(-5)

You are moving down the numberline, then subtracting again because of the negative sign. -(1+5) is how I like to think of it.

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2 years ago

The amoebas on a petri dish is doubling every 5 minutes. If there are currently 8 amoebas on the dish, how many will there be in

pychu [463]

Answer:

32,768.

Step-by-step explanation:

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2 years ago

The ratio of girls to students at a middle school is 10 to 18. If there are 270 students at the school, how many are girls?

Stells [14]

A good way of doing this problem would be to use cross multiplication of ratios
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Then cross multiply

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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

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