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.230769231

get a calculator

   the first question is 2 pets

the answer is 5 because if you squared the 5 the result is 25. Squared means multiplying two times to the original number.

Thank you.

T tried 1st problem. Answer i got is A . 1 pet

is it correct ?

Area of cross-section times length.  For a more specific answer you'd need to specify the shape of the wedge (with some values or parameters describing its various dimensions). 

I'm afraid neither your question nor your answer makes any mathematical sense.  What do you mean by log^d? Do you just mean log(d)?  Do you mean log(d/10)? Do you mean log(d)/10? Do you mean something else?  

And just because your log might be to base 10, that doesn't mean you can just cancel it with another 10.  

No thanks to anyone else, I think I've got the answer.      1.08=log^d/10

This is a log base 10. That means I can cancel out the 10's which leaves me with 1.08=logd.

Now I just need to rewrite this as an exponential equation. 10^1.08=d.

d is approximately 12.

The surface area of a cylinder is given by 2*pi*r*h

Using 3.14 for "pi,'  we have 2*(3.14)*(3m)*(30m) = 565.2 m²

But we only want 1/2 of this........282.6m²

Looks like "a" is your answer.

How do you solve 7-9x-(x+12) < 25?

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Let's start by simplifying the left side - note that the "-" outside the parenthesis makes both the "x" and the "12" inside negative !!    We have

-10x - 5  < 25       add 5 to both sides

-10x < 30        divide by -10 on both sides and reverse the inequality sign (as we always do when multiplying or dividing an inequality by a negative)

x > -3

The square root of 25 is 5 because 5x5 is 25.

Q#1-A You want to list out the number of pets per people in the class. If 8 people have 0 pets, list 0 eight times. List the number 1, six times. List the number 2, three times. List the number 3, three times. So your list of numbers would be 00000000111111222333. Cross off the first number, then cross off the last number. Continue doing this until there is one number left in the middle. That will be your answer.

Q#2-D Repeat the process with the given information.

Q#3-A Repeat the process with the given information.

1. Combine like terms.

-9x combines with -x to make -10x

7 combines with 12 to make 19

Your equation should look like this now: -10x+19 < 25

2. Solve for x

Subtract 19 from both sides

At this point your equation should look like this -10x < 6

Then, you divide both sides by -10

Then you'll get x > -6/10 (the inequality sign switches because you divide by a negative)

3. Simplify

x > -6/10 simplifies to x > -3/5

x > -3/5 is your answer! 

It's a graph of a linear function with a y-intercept at (0, 7) and an x-intercept at (7, 0) that looks like this:

No, but 2π radians make a complete circle (i.e. is the equivalent of 360°).  One radian is the angle subtended at the centre of the circle by an arc of the circle such that the arc has the same length as the radius of the circle.  One radian is 360/2π degrees:

$${\mathtt{radian}} = {\frac{{\mathtt{360}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}} \Rightarrow {\mathtt{radian}} = {\mathtt{57.295\: \!779\: \!513\: \!082\: \!320\: \!9}}$$ degrees

 Note: My "No" refers to the question, not Melody's answer! Melody must have posted while I was writing.

No $$2\pi$$ is an irrational number.

But

radians can be used to measure angles and $$2\pi$$ radians is equal to 360 degrees.   

does that make sense?  

The tangent is opposite over adjacent, so if we let opposite be 8 and adjacent be 15 then the hypotenuse is 

$${\sqrt{{{\mathtt{8}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{15}}}^{{\mathtt{2}}}}} = {\mathtt{17}}$$

cosine is adjacent over hypotenuse, so cos(B) = 15/17.  Angle A must be the other angle (not the right-angle) so we must have cos(A) = 8/17.

Even if the sides are not actually 8, 15 and 17 they must all be a scale factor times those values.  The scale factor will cancel out of the ratios when determining cos, sin and tan.

1,000,000/2401

Deleted. I misread the question. 

exponents (3/5)^-8/(5/3)^-6-3^-11(1/3)^-9

$${\frac{{\left({\frac{{\mathtt{3}}}{{\mathtt{5}}}}\right)}^{\left(-{\mathtt{8}}\right)}}{{\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right)}^{\left(-{\mathtt{6}}\right)}}}{\mathtt{\,-\,}}{{\mathtt{3}}}^{\left(-{\mathtt{11}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}^{\left(-{\mathtt{9}}\right)} = {\mathtt{1\,275.982\: \!383\: \!327\: \!176\: \!069\: \!9}}$$

Equations like this need a numerical (iterative) method for their solution.

First though is to sketch a graph or graphs in order, if possible, to determine the number and approximate locations of the roots. For this particular example, best is to rewrite the equation as x² - 4x = 2^x and to sketch the graphs of y = x² - 4x, which is a concave up parabola crossing the x-axis at x=0 and x=4, and y = 2^x which looks rather like the y = e^x curve. At their point(s) of intersection x² - 4x = 2^x in which case their x co-ordinates will be the solutions to the original equation. There is just one intersection to the left of the origin, located (after trying two or three values for x), round about x = -0.2, and no intersections to the right, (the 2^x curve being already at a height of 16 as the parabola becomes positive at x=4).

The most popular numerical method for solving non-linear equations is probably the Newton-Raphson method.

For this equation the formula is

$$x_{n+1}=x_{n}-\frac{x^{2}_{n}-4x_{n}-2^{x_{n}}}{2x_{n}-4-2^{x_{n}}\ln(2)}$$

which, starting with x₀ = -0.2 converges to 0.206100 after just two iterations.

It's also usually possible to obtain a convergent iterative formula from the original equation. Rewriting the original equation as x = (x² - 2^x)/4 and using this as an iterative formula, (substitute x=-0.2 on the rhs to get x=-0.2076, substitute this into the rhs etc.), produces the sequence -0.2, -0.2076, -0.2057, -0.2062, -0.20608, etc. Not as fast as Newton-Raphson but it will get there eventually.

This may be in Croatian and this might be the question.

How to calculate how much is half of the forth root of 10

$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)} = {\mathtt{0.889\: \!139\: \!705\: \!019\: \!461\: \!4}}$$

You have to get past 1 + 1, first.

That might overload your brain.

Was this supposed to have a picture with it?  It doesn't make much sense.  Sorry.

Thanks Alan for the matrix info.

With regards to the page numbering, yes, I think you are missing something.

It is very obscure.  sometimes individual post in a thread can be referenced but I don't understand it.  I have deleted the duplicates now but in my temporary sticky questions today I referenced threads and they doubled up even though the numbers were different.  I have known this is the case for some time but as I said, I do not understand how it works or how i can make proper use of it.

Oh, on the answer page, individual posts, not threads, are referenced all the time.

Also if you do a search - which isn't working very well yet - often a post will come up but it is stand alone, the thread it belongs to is nowhere in sight!

@@ End of Day Wrap - Sunday 4/5/14   Sydney, Australia Time 19:20

Hi everyone,

Our wonderfully impressive answers were posted by CPhill, Problem, Bertie, Alan, Kitty

I had real problems getting into this thread.  I kept being told "File not found" I guess there is still quite a few ants at the picnic!  

Another recurring complaint is that time stamps on out messages jump all over the "clock" 

Andre has now installed a fraction calculator.  There is a drop down menu in the top right corner of the calculator.  There is a few different calculators available here.