Math:Kimberly took her 6 nieces and
nephews to a hockey game. She
wants to buy them snacks. How
much can she spend on snacks for
each child if Kimberly wants to
spend less than
$33
in total

Annette Arroyo
2020-11-10
Answered

You can still ask an expert for help

sweererlirumeX

Answered 2020-11-11
Author has **91** answers

There are 66 children and in all 33 dollars.
So, she can spend at most $d\frac{33}{6}=d\frac{11}{2}=5.5$ dollars on each child.
Thus, the answer is $ 5.50, or if it must be an integer $ 55.

asked 2021-05-11

Evaluate the line integral, where C is the given curve

C xy ds

C:$x={t}^{2},y=2t,0\le t\le 5$

C xy ds

C:

asked 2021-02-24

asked 2021-08-11

Find an equation of the following curve in polar coordinates and describe the curve.

asked 2022-07-18

Find all asymptotes of a function

Find all asymptotes of a function:

$f(x)=\mathrm{log}({x}^{2}-4)$

Domain: $x\in (-\mathrm{\infty},-2)\cup (2,\mathrm{\infty})$

Vertical asymptotes are x=−2 (left) and x=2(right):

$\underset{x\to \text{}-{2}^{-}}{lim}\mathrm{log}({x}^{2}-4)=-\mathrm{\infty}$

$\underset{x\to \text{}{2}^{+}}{lim}\mathrm{log}({x}^{2}-4)=-\mathrm{\infty}$

I calculate the limits in +/- infinity:

$\underset{x\to \text{}+\mathrm{\infty}}{lim}\mathrm{log}({x}^{2}-4)=+\mathrm{\infty}$

$\underset{x\to \text{}-\mathrm{\infty}}{lim}\mathrm{log}({x}^{2}-4)=+\mathrm{\infty}$

So I'm looking for the oblique asymptotes of a form y=Ax+B:

${A}_{+}=\underset{x\to \text{}+\mathrm{\infty}}{lim}\frac{\mathrm{log}({x}^{2}-4)}{x}=\underset{x\to \text{}+\mathrm{\infty}}{lim}\frac{\frac{2x}{{x}^{2}-4}}{1}=0$

${B}_{+}=\underset{x\to \text{}+\mathrm{\infty}}{lim}\mathrm{log}({x}^{2}-4)-{A}_{+}x=+\mathrm{\infty}$

The same for $-\mathrm{\infty}$. How should I interpret this? There are no oblique asymptotes?

Find all asymptotes of a function:

$f(x)=\mathrm{log}({x}^{2}-4)$

Domain: $x\in (-\mathrm{\infty},-2)\cup (2,\mathrm{\infty})$

Vertical asymptotes are x=−2 (left) and x=2(right):

$\underset{x\to \text{}-{2}^{-}}{lim}\mathrm{log}({x}^{2}-4)=-\mathrm{\infty}$

$\underset{x\to \text{}{2}^{+}}{lim}\mathrm{log}({x}^{2}-4)=-\mathrm{\infty}$

I calculate the limits in +/- infinity:

$\underset{x\to \text{}+\mathrm{\infty}}{lim}\mathrm{log}({x}^{2}-4)=+\mathrm{\infty}$

$\underset{x\to \text{}-\mathrm{\infty}}{lim}\mathrm{log}({x}^{2}-4)=+\mathrm{\infty}$

So I'm looking for the oblique asymptotes of a form y=Ax+B:

${A}_{+}=\underset{x\to \text{}+\mathrm{\infty}}{lim}\frac{\mathrm{log}({x}^{2}-4)}{x}=\underset{x\to \text{}+\mathrm{\infty}}{lim}\frac{\frac{2x}{{x}^{2}-4}}{1}=0$

${B}_{+}=\underset{x\to \text{}+\mathrm{\infty}}{lim}\mathrm{log}({x}^{2}-4)-{A}_{+}x=+\mathrm{\infty}$

The same for $-\mathrm{\infty}$. How should I interpret this? There are no oblique asymptotes?

asked 2021-11-11

Evaluate the improper integral using limits.

$\int}_{1}^{2}\frac{dx}{\sqrt{3x-2}$

asked 2022-07-18

Concavity and critical numbers

I am attempting to study for a test, but I forgot how to do all the stuff from earlier in the chapter.

I am attempting to find the intervals where f is increasing and decreasing, min and max values and intervals of concavity and inflection points.

I have $f(x)={e}^{2x}+{e}^{-x}$. I know that I have to find the derivative, set to zero to find the critical numbers then test points then find the second derivative and do the same to find the concavity but I don't know how to get the derivative of this. Wolfram Alpha gives me a completely different number than what I get.

I am attempting to study for a test, but I forgot how to do all the stuff from earlier in the chapter.

I am attempting to find the intervals where f is increasing and decreasing, min and max values and intervals of concavity and inflection points.

I have $f(x)={e}^{2x}+{e}^{-x}$. I know that I have to find the derivative, set to zero to find the critical numbers then test points then find the second derivative and do the same to find the concavity but I don't know how to get the derivative of this. Wolfram Alpha gives me a completely different number than what I get.

asked 2022-07-18

Question on Local Maxima and Local Minima

Find the set of all the possible values of a for which the function $f(x)=5+(a-2)x+(a-1){x}^{2}-{x}^{3}$ has a local minimum value at some $x<1$ and local maximum value at some $x>1$.

The first derivative of f(x) is:

${f}^{\prime}(x)=(a-2)+2x(a-1)-3{x}^{2}$

I do know the first derivative test for local maxima and local minima, but I can't figure out how I could use monotonicity to find intervals of increase and decrease of f′(x)

The expression for f′(x) might suggest the double derivative test is the key, considering ${f}^{\u2033}(x)=2(a-1)-6x$ for which the intervals where it is greater than zero and less than zero can be easily found, but then again I can't think of a way how I could find a c such that ${f}^{\prime}(c)=0$.

Find the set of all the possible values of a for which the function $f(x)=5+(a-2)x+(a-1){x}^{2}-{x}^{3}$ has a local minimum value at some $x<1$ and local maximum value at some $x>1$.

The first derivative of f(x) is:

${f}^{\prime}(x)=(a-2)+2x(a-1)-3{x}^{2}$

I do know the first derivative test for local maxima and local minima, but I can't figure out how I could use monotonicity to find intervals of increase and decrease of f′(x)

The expression for f′(x) might suggest the double derivative test is the key, considering ${f}^{\u2033}(x)=2(a-1)-6x$ for which the intervals where it is greater than zero and less than zero can be easily found, but then again I can't think of a way how I could find a c such that ${f}^{\prime}(c)=0$.