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3 years ago

Find the solution set for this inequality below.

Mathematics

1 answer:

Maksim231197 [3]3 years ago

3 0

Answer:

x < -1

Step-by-step explanation:

-3x - 4 > 8x + 7

-3x -8x > 7 + 4

-11x > 11 / : (-11) **

x < -1

** when dividing by a negative number, the sign changes

If you have any questions about the way I solved it, don't hesitate to ask me in the comments below :)

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Step-by-step explanation:

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2 years ago

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SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

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3 years ago

What is the simplified form of the expression 3a + 7 - a - 7

Sonja [21]

You can simplify expressions by simplifying parentheses by using the distributive property and combing like terms.

Combining like terms is the process of adding up numbers with the same variable(s) and exponent. If a term does not have an exponent then an exponent of 1 is implied.

There is no parentheses in the expression 3a + 7 - a - 7
But, there are like terms in the expression.

Which are the like terms?

3a and a are both like terms. They both have the same variables and they both have the same exponent.

7 and -7 are like terms, too.

3a + 7 - a - 7
2a

So, 2a is the simplified version.

6 0

3 years ago

Read 2 more answers

Write the expression nine divided by the quantity of x plus y

seropon [69]

9514 1404 393

Answer:

$\text{A)}\quad\dfrac{9}{x+y}$

Step-by-step explanation:

In the English expression, you need to understand that "A divided by B" means that A is the numerator and B is the denominator of the fraction representing the division. Here, "B" is "the quantity of ...", which means that the stuff after "of ..." is to be treated as a single item, the denominator in this case. That item is "x plus y." So, the expression in math terms is ...

$\dfrac{9}{x+y}$

When this is written in plain text, the "quantity" is put in parentheses, because the fraction bar no longer serves as a grouping symbol:

9/(x+y)

These expressions match the first choice, A.

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OleMash [197]

Answer:

Step-by-step explanation:

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