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3 years ago

-2(3x+4)<-4+8 solve the inequality

Mathematics

2 answers:

Shtirlitz [24]3 years ago

8 0

Answer:

Inequality: x > -2

Interval Notation: ( -2, ∞)

Step-by-step explanation:

mario62 [17]3 years ago

8 0

Answer:

x > -2

Step-by-step explanation:

-2(3x+4) < -4+8 (evaluate right side)

-2(3x+4) < 4 (expand the parentheses on the left side using distributive property)

3x(-2) + 4(-2) < 4

-6x - 8 < 4 (add 8 to both sides)

-6x < 4 + 8

-6x < 12 (divide both sides by 6)

-x < (12/6)

-x < 2 (multiply both sides by -1; remember to flip the sign when multiplying both sides by a negative number)

x > -2

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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 $y = e^x$

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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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2 years ago

What is a 96% confidence interval for the proportion of weight-lifting injuries in this age group that were accidental?

loris [4]

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2 years ago

What is the value of x?

musickatia [10]

X = 7. Because both lines equal the same so it has to equal 75
11 x 7 = 77
77-2 = 75
X = 7

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2 years ago

Read 2 more answers

Please what is the answer?

nirvana33 [79]

Answer:

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