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romanna [79]
3 years ago
5

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>

3 0
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.

Since . z=\dfrac{x-\mu}{\sigma}

For x= 1030 , we have

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

Hence, the required z-score = -1.7

5 0
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:

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]\\\\

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
Please help! Picture Down Below!
guajiro [1.7K]

Angle 4 = Angle 5 = 42⁰( alternate interrior angle)

Hope it helps u

5 0
3 years ago
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