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3 years ago

Simply 8x squared + 5(7x squared- x) - 2x =

Mathematics

1 answer:

Slav-nsk [51]3 years ago

4 0

Suuuuree,

$8 {x}^{2} + 5(7 {x}^{2} - x) - 2x \\ \\ 8 {x}^{2} + 35 {x}^{2} - 5x - 2x \\ \\ 43 {x}^{2} - 7x$

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The length width and height of a rectangular solid are 8,4,1 respectively what is the length of the longest line segment whose e

LUCKY_DIMON [66]

Answer:

9 units

Step-by-step explanation:

Given

$Length = 8$

$Width = 4$

$Height = 1$

Required

Determine the length of the longest segment

To do this, we use:

$Longest = \sqrt{Length^2 + Width^2 + Height^2}$

So, we have:

$Longest = \sqrt{8^2 + 4^2 + 1^2}$

$Longest = \sqrt{64 + 16 + 1}$

$Longest = \sqrt{81}$

Take positive square root

$Longest = 9$

<em>Hence, the length of the longest segment is 9 units</em>

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3 years ago

Enter the answer to the nearest tenth. Another home business has an average income of $1.200.00 per month with a standard deviat

Nonamiya [84]

Answer: -1.7

Step-by-step explanation:

Given : Another home business has an average income of $1.200.00 per month with a standard deviation of $100.00.

i.e. $\mu=\$1200.0$

$\sigma=\$100.00$

Let x represents the monthly income.

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For x= 1030 , we have

$z=\dfrac{1030-1200}{100}=-1.7$

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3 years ago

A company makes a profit of $2 per software program and $3 per video game. The company can produce at most 40 software programs

Helen [10]

Answer:

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Step-by-step explanation:

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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

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