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3 years ago

6 - 4 (6n + 7 ) > 122

Mathematics

1 answer:

Galina-37 [17]3 years ago

3 0

Answer:

n<-6

Step-by-step explanation:

Multiple parentheses by 4
Calculate it
Move constant to the right
Add the numbers
Divide both sides by -24

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In 2000, U.S. trade with Saudi Arabia totaled $20.6 billion. Of this total,

xxMikexx [17]

Answer:

30.1%

Step-by-step explanation:

Find the percent that was US exports by dividing 6.2 by 20.6, then multiplying it by 100

(6.2/20.6) x 100

= approximately 30.1

So, approximately 30.1% of the trade was US exports

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3 years ago

Select the correct answer.

gizmo_the_mogwai [7]

Answer:

Factors are (x-1), (x+1), and (x+2).

Step-by-step explanation:

$g(x)=x^{3}+2x^{2} -x-2\\g(x)=(x - 1) (x + 1) (x + 2)$

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3 years ago

What’s the answer to this question. <br> Simple correct answers

Sophie [7]

Answer:

1st one

Step-by-step explanation:

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4 years ago

Solve the equation.<br> 4+ 14 = – 2(7x-9)

Tamiku [17]

Answer:

-63

Step-by-step explanation:

4 0

3 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

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.

6 0

3 years ago