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3 years ago

Find the quotient of 90 over -10

Mathematics

2 answers:

OLEGan [10]3 years ago

5 0

Hi! The answer is -9 because 90/-10 = -9
I hope this helps you, Goodluck! :)

vodka [1.7K]3 years ago

3 0

90/-10

= 9/-1

= -9

So, -9 is the quotient.

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(6+8)^2=6^2+8^2 statement true

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If there are 24 cats in the exhibit, how many tigers do you have?

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3 years ago

Read 2 more answers

How many solutions are there to the equation below?<br>12x + 32 = 12X-7

Greeley [361]

Answer: no solutions

Step-by-step explanation:

12x+32=12x-7

12x-12x=-7-32

0x=-39

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3 years ago

A 13 ft ladder is resting on a wall. The angle the ladder makes with the floor is 45 degrees. How high off the ground is the top

PtichkaEL [24]

Answer: 9.19 ft

Step-by-step explanation:

Hi, since the situation forms a right triangle (see attachment) we have to apply the next trigonometric function.

Sin α = opposite side / hypotenuse

Where α is the angle of elevation of the ladder to the ground, the hypotenuse is the longest side of the triangle (in this case is the length of the ladder), and the opposite side (x) is distance between the top of the ladder and the ground.

Replacing with the values given:

Sin 45 = x/ 13

Solving for x

sin45 (13) =x

x= 9.19 ft

Feel free to ask for more if needed or if you did not understand something.

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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

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

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

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2 years ago