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Find the smallest distance between the origin and a point on the parabola y = x^2 - 2.

(where I have distance from origin, I should have put distance squared from origin!)

Sometimes accounts aren't logged in. 

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Thanks 

See https://web2.0calc.com/questions/math_83467#r2

that would be 30

Number of CDs = 2*3*5 + 1

NP at least you saw this question 

Is this asinus?

nvm it is why does it show him as guest?

20 umbrellas

20/2 = 10 red

5 blue 

20 - 10 - 5 = 5 black

:))

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Hello :))

I'm a bit confused on how f(x) = 192/x +16x + 24 went to the second part. 

What does d stand for?

Thanks

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Hey there, ellapow!

So let's solve this problem.

\(\left(-1\right)^2+5\left(-1\right)-6\)

\(1+(5)(−1)−6\)

\(1+−5−6\)

\(−4−6\)

\(x=-10\)

Therefore, x equals -10.

Hope this helped! :)

( ﾟдﾟ)つ Bye

-30 = (x)(7-x)

-30 = 7x - x^2

x^2 - 7x - 30 = 0

(x-10)(x+3) = 0

Our answers are 10, and -3.  :))

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She has an ON OFF OFF pattern. We want to maximize OFF days. Let's start the pattern on day zero (Jan 31st).  So now she is going OFF OFF ON.

Since that's a 3 day rotation, we can say that every FULL 3 day segment is one walk. 28/3 = 9.333. Drop the decimal since we just proved it's an off day, you get 9.

28 days in February. 

Every third day is 28/3 = 9.333 = 9 days of walking. 

9*4 = 36

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All the points that will be equidistant from two parallel lines will also have the same slope, and be in the center of those lines. :))

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Thank you for creating the graph, it helped me visualize the angles and line segment ratios better than my messy drawing. :))

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Thank you for answering. :))

I understand now. 

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Think about this incrementally, adding one shape at a time.

1. Circle, zero intersection

2. Circle, best we can do is 2, making a venn diagram.

3. Line, let's have it be horizontal and cut the venn in half. That's 4 more, so we're at 6.

4. Line, this one can be almost the same as the last one but just a tiny bit off horizontal. It cuts the circles for 4 more intersections, and the existing line for 1 more.

That's 11 total. If we think about each shape individually, this is the best possible result

I graphed it to scale and also got 40 degrees.

180(n-2)=1070+m     where m<180  and   (1070+m)mod180=0

n-2=(1070+m)/180

n-2=5+(170+m)/180

m=10

n-2=5+(170+10)/180

n-2=5+1

n-2=6

n=8

This should help you answer:

A rectangular container with open top is required to have a volume of 24 cubic meters. Also, one side of the rectangular base is required to be 4 meters long. If material for the base costs \(8\ per\ square\ meter,\ and\ material\ for\ the\ side’s\ costs \) 2 per square meter, find the dimensions of the container so that the cost of material to make it will be a minimum.

Hello Guest!

V = abh = 24

h = x

a = 4

4bx = 24

b = 6/x

\(Costs = f(x)=ab\times 8 +2(ax+bx)\times 2\\ f(x)=4\cdot\frac{6}{x}\cdot 8 +2(4x+\frac{6}{x}\cdot x)\cdot 2\\ x\in \mathbb R|x=0\\ f(x)=\frac{192}{x}+16x+24\)

\(\frac{df(x)}{dx}=-\frac{192}{x^2}+16=0\\ 16x^2=192\\ x^2=12\\ \color{blue}x=3.464\\\)

When the cost is minimal, the rectangular container has a height of 3.464.

You can do the rest of the task yourself.

  !

asinus

Find the number of terms in the sequence.
\(-18,~ 36,~ -72,~\ldots ~,~9437184\)

My attempt:

\(\begin{array}{|r|rl|} \hline \text{sequence} & : 9 \\ \hline -18 & -2 & = (-2)^1 \\ \hline 36 & 4 & = 2^2 \\ \hline -72 & -8 & = (-2)^3 \\ \hline \ldots & \ldots & \ldots \\ \hline 9437184 & 1048576 & = 2^{20} \\ \hline \end{array}\)

\(\text{Formula:}\\ \begin{array}{|rcll|} \hline \mathbf{a_n} &=& \mathbf{9*(-2)^n} \\ a_{20} &=& 9*(-2)^{20} \\ a_{20} &=& 9*2^{20} \\ a_{20} &=& 9*1048576 \\ \mathbf{a_{20}} &=& \mathbf{9437184} \\ \hline \end{array}\)

There are 20 terms in the sequence.

It was very well authored and easy to understand. Unlike additional blogs I have read which are really not that good.