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For the first two, you can use Pythagorean Theorem. This states that \({a}^{2}+{b}^{2}={c}^{2}\), where \(a\) and \(b\) are the legs, and \(c\) is the hypotenuse. 

Problem 1: We already have sides \(a = 10\) and \(c = \sqrt{170}\). We can move the formula around for it to fit this problem better. We can change it to \({b}^{2}={c}^{2}-{a}^{2}\). When subsituting our values in, we can get \({b}^{2}={\sqrt{170}}^{2}-{10}^{2}\). When we square both values, we can get \({b}^{2} = 170-100\), so \({b}^{2} = 70\). We can solve this to be \(b = \sqrt{70}\) or \(x = \sqrt{70}\).

Problem 2: We already have sides \(b = 5\sqrt{2}\) and \(c = 13\). We can move the formula around for it to fit this problem better. We can change it to \({a}^{2}={c}^{2}-{b}^{2}\). When plugging in our values, we can get \({a}^{2} = {13}^{2} - ({5\sqrt{2})}^{2}\). By solving this, we can get \({a}^{2} = 119\) which equals \(a = \sqrt{119}\) or \(x = \sqrt{119}\).

- Daisy

From these two sentences, we can get two equations: \(3s + 7t = 134\) and \(3s + 4t = 92\). In order to solve this, we can use elimination, to get rid of one of the variables. If we subtract the second equation from the first, we can get \(3t = 42\), which simplifies to \(t = 14\). Now that we know that the student tickets cost $14, we can plug that into either the first or second equation. Both will give the same answer. Let's just use the first equation. From here we can get \(3s + 7(14) = 134\), which we can solve and get \(s = 12\). So the senior tickets cost $12.

- Daisy

Why Mr. BB, do you mean to tell me my answer is ………..BLARNEY!!

Blarney answers are acceptable for blarney questions, Mr. BB. It doesn’t matter how absurd the reasoning is, as long as the math is correct, and it is: The weighted harmonic mean is mathematically correct, and the method used to weight it is mathematically correct. The logic is correct in another universe (not this one), but it’s usable here because this question, and the obsequiously (means ass-kisser) polite student who asked it, came from came from the same universe.

In this world, a competent teacher has countless questions and scenarios for assigning practice and to assess a student’s TVM skills. A teacher would only present this question to demonstrate absurdity. No one except a fool would consolidate a loan using this approach, and only an unethical loan company would make such a loan. I can think of one that would: The First National Vigorish and Loan-Shark Company; a wholly-owned subsidiary of Smith, Blarney, Cheatem, and Howl, your former and principal employer.

The absurdity becomes very clear when you add another year of payments (vigorish) to a fully amortized loan.

Another thing that becomes clear is you are the Blarney Banker. … … …

GA

I have been asked if I can try and explain a little more.

\(\text{Sketch the graph }\quad y= \frac{\pi}{2}\; sin^2x \quad \text{and use it to solve the equation }\quad2x=\pi\;sin^2x\)

\(2x=\pi\;sin^2x\\ \text{rearranging we get}\\ x=\frac{\pi}{2}sin^2x\)

\(\text{So we have}\quad y= \frac{\pi}{2}\; sin^2x \quad\\ \text{and we want to use this graph to determine when } \quad x= \frac{\pi}{2}\; sin^2x \\ \text{but we have the graph of } \quad y= \frac{\pi}{2}\; sin^2x \quad \\ \text{So we want to know when x=y on this graph.}\\ \text{i.e. }\quad y=x=\frac{\pi}{2}\; sin^2x\\ \text{So if we find the intersection of } \qquad y=\frac{\pi}{2}\; sin^2x \qquad and \qquad y=x\\ \text{That will be all the solutions of }\qquad x= \frac{\pi}{2}\; sin^2x \)

\(\text{The intersection happens when }\quad    x= 0, \;\frac{\pi}{4} \;\;and  \;\frac{\pi}{2}         \\ \text{These are the only solutions to  }  x=\frac{\pi}{2}sin^2x\\\)

I hope that helps more   

Look at Melody's answer here: https://web2.0calc.com/questions/help_94831#r3 

\(\text{Sketch the graph }\quad y= \frac{\pi}{2}\; sin^2x \quad \text{and use it to solve the equation }\quad2x=\pi\;sin^2x\)

I have been asked how to plot the original graph.

I would 'build' this graph.  Starting with y=sinx

This has a period of 2pi and a amplitude of 1

So you can graph this.

\( \text{Not that }\quad2x=\pi\;sin^2x \quad \text{is the same as } x=\frac{\pi}{2}\;sin^2x\)

So you want the intersection of     \(y=\frac{\pi}{2}\;sin^2x\)    with   \(y=x\)

Draw the sine graph and the graph of y=x and it is very easy to see where they intersect.

I hope that help  

Hi,

I struggled a little to get my head around it too.   

I'm letting y=f(x) some of the time because it is easier for me to work with.

there is a vertical asymptote at x=3 this means that  y tends to =+ or - infinty when x=3

This will happen if the denominator tends to 0 as x tends to 3

hence c=-3

There is a horizontal asymptote at y=-4.

This means that as y tends to  -4,     x  tends to =+ or - infinty 

So 

  \(\displaystyle \lim_{x\rightarrow \infty}\; f(x)=-4\qquad or \qquad \displaystyle \lim_{x\rightarrow -\infty}\; f(x)=-4\\ a=-4\qquad or \qquad a=-4\\ so\;\;a=-4 \)

so we have

\(f(x) = \frac{-4x+b}{x-3}\)

Now sub in  ( 1,0) to get the value of b 

you will get b=4

\(f(x) = \frac{-4x+4}{x-3}\)

If you think about the asymptotes on a graph it might help. It did help me.

And here is the actual graph

\(\text{Sketch the graph  }\quad y= \frac{\pi}{2}\; sin^2x  \quad  \text{and use it to solve the equation  }\quad2x=\pi\;sin^2x\)

Haha I'm fine, thank you very much, though! I'll keep that in mind the next time I have trouble on a question.

- Daisy

Don't apologize! I give my time freely. If I help someone learn and get a thank you then it makes me very happy.  :))

If you still are having trouble filling in the blanks I am quite happy to elaborate :)  You only have to ask :)

Good, thank you :)

Haha I didn't really answer the question, but yeah I agree with you.

I am sorry melody. I will not delete my questions in the future.

Max that sounds like total nonsense.

If your teacher did not want other kids to cheat they would not want you to cheat either!

You take if off to attempt to disguise the fact that you are cheating.

It is extremely rude. People have been banned from this site for doing this.

People go to the effort of answering your question and then you blank your question. You have not even bothered to give Olpers a proper thankyou.

You have not even given him/her a point although I notice you are quite happy to give yourself a point just for asking a question.

It is not good that you have drawn attention to yourself in this way.

Do not ever delete any of your question in the future!

It's well drawn, and it helped me comprehend the solution. Thanks!

- Daisy

total nonsense xD

Thanks! That answered my question about the diagram! Sorry for taking up your time!

- Daisy

Here is my diagram

I get the same answer as Melody....but....I  don't know if the included pic is accurate or not

I have assumed that  angle ABC  =  90 =  angle BAD

Since AC  bisects  angle BAD, then angle BAC =  angle CAD  = 45

AB = BC

So... AC  = √2AB  = CD

Area   of ABCD   =  Area  of triangle ABC   + Area of triangle ACD  = 42

So

Area  of triangle ABC  + area of triangle ACD  = 42

(1/2)AB*BC   + (1/2)AC * CD  = 42

(1/2)AB^2  + (1/2)√2AB * √2AB   = 42

AB^2  + 2AB^2  = 84

3AB^2  = 84

AB^2 = 28  = area of triangle ACD

Haha why did you delete the question?

um...ok well thanks i guess

Thank you! The picture is very clear!

Problem \(1\):

The formula for the surface area of a cone is \(πrl+π{r}^{2}\), where \(r\) is the radius and \(l\) is the slant height. For problem one, \(r = 6\) and \(l = 14\). By plugging in the variable values, we can get \(π(6)(14)+π{6}^{2} = 84π + 36π = 120π\). 

Problem 2:

The formula for the surface area of a cylinder is \(2πrh+2π{r}^{2}\), where \(r\) is the radius and \(h\) is the height. \(r = 13\) and \(h = 17\). When you plug in the variable values, you can get \(2π(13)(17)+2π{13}^{2} = 442π + 338π = 780π\), and it is asking for the nearest whole number, so multiply \(3.14\) by \(780\). The product is \(2449.2\), which rounds down to \(2449\).

Problem 3:

The formula for the surface area of a cone is \(πrl+π{r}^{2}\), where \(r\) is the radius and \(l\) is the slant height. \(r = 7\) and \(l = 24\). By plugging in the variable values, we can get \(π(7)(24)+π{7}^{2} = 168π + 49π = 217π\). Since they are asking for a decimal to the nearest tenth, we can get \(681.4\).

Problem 4:

The formula for the surface area of a cylinder is \(2πrh+2π{r}^{2}\), where \(r\) is the radius and \(h\) is the height.  \(r = 25\) and \(h = 36\). When you plug in the variable values, you can get \(2π(25)(36)+2π{25}^{2} = 1800π + 1250π = 3050π\).

- Daisy