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3 years ago

Simplify (8j3 + 5j2 – 3) – (5j3 + 7j2 – 12j + 7) ...?

2 answers:

5 0

The answer to the problem is:

8j^3 + 5j^2 - 3 - 5j^3 - 7j^2 + 12j - 7

3j^3 - 2j^2 + 12j - 10


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

Over [174]3 years ago

4 0

Answer:

the answer is C

Step-by-step explanation:

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4 years ago

What is 10*60,000 equal to

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4 years ago

Assume that adults have IQ scores that are normally distributed with a mean of mu equals 105 and a standard deviation sigma equa

kondor19780726 [428]

Answer: 0.6827

Step-by-step explanation:

Given : Mean IQ score : $\mu=105$

Standard deviation : $\sigma=15$

We assume that adults have IQ scores that are normally distributed .

Let x be the random variable that represents the IQ score of adults .

z-score : $z=\dfrac{x-\mu}{\sigma}$

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For x= 120

$z=\dfrac{120-105}{15}\approx1$

By using the standard normal distribution table , we have

The p-value : $P(90$

$P(z$

Hence, the probability that a randomly selected adult has an IQ between 90 and 120 =0.6827

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3 years ago

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NEED HELP ASAP giving brainly

uranmaximum [27]

Answer:

A&C

Step-by-step explanation:

They are negative and even and they are not positive and even too.

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3 years ago

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Use the substitution x = et to transform the given Cauchy-Euler equation to a differential equation with constant coefficients.

Anika [276]

Answer:

$\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \ }$

Step-by-step explanation:

Hello,

let's follow the advise and proceed with the substitution

first estimate y'(x) and y''(x) in function of y'(t), y''(t) and t

$x(t)=e^t\\\dfrac{dx}{dt}=e^t\\y'(t)=\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=e^ty'(x)y'(x)=e^{-t}y'(t)\\y''(x)=\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(e^{-t}\dfrac{dy}{dt})=-e^{-t}\dfrac{dt}{dx}\dfrac{dy}{dt}+e^{-t}\dfrac{d}{dx}(\dfrac{dy}{dt})\\=-e^{-t}e^{-t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}=-e^{-2t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}e^{-t}\\=e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt})$

Now we can substitute in the equation

$x^2y''(x)+9xy'(x)-20y(x)=0\\ e^{2t}[ \ e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}) \ ] + 9e^t [ \ e^{-t}\dfrac{dy}{dt} \ ] -20y=0\\ \dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+ 9\dfrac{dy}{dt}-20y=0\\ \dfrac{d^2y}{dt^2}+ 8\dfrac{dy}{dt}-20y=0\\$