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2 years ago

Find the solution set of the inequality \qquad52- 3x < -14.52−3x<−14.

Mathematics

2 answers:

SIZIF [17.4K]2 years ago

7 0

Answer:

52 - 3x < - 14.52 − 3x < − 14 = x ∈ ∅

Step-by-step explanation:

- 14.52 - 3x > 52 - 3x

- 14.52 - 3x < - 14

- 14.52 > 52

- 3x < - 14 + 14.52

x ∈ ∅

- 3x < 0.52

x ∈ ∅

x > - $\frac{13}{75}$

x ∈ ∅

I think that's true

Anarel [89]2 years ago

5 0

Answer:

Step-by-step explanation:

add 14.52 to both sides of the inequality to cancel that value out from the middle. Then you can divide both sides by 3x to get x by itself in the middle.

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Answer:

-21

Step-by-step explanation:

3t-7=5t

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3 years ago

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Nady [450]

Answer:

Step-by-step explanation:

a. 1 & 2

b. 3 & 5

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2 years ago

A ball will be drawn from the bag containing 20 balls numbered one through 20 if each ball is equally likely to be drawn what is

son4ous [18]

Answer:

3/5

Step-by-step explanation:

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Of these, 10 are even, 4 are less than five, and 2 are both even and less than 5.

So the probability is:

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2 years ago

Use the definition of a Taylor series to find the first three non zero terms of the Taylor series for the given function centere

Ket [755]

Answer:

$e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...$

$\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n$

Step-by-step explanation:

Taylor series expansions of f(x) at the point x = a

$\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...$

This expansion is valid only if $\text{f}\:^{(n)}(a)$ exists and is finite for all $n \in \mathbb{N}$ , and for values of x for which the infinite series converges.

$\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1$

$\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...$

$\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}$