power27

2. Call the original fraction \(\frac{n}{d}\), and the modified fraction \(\frac{n+5}{d+5}=\frac{1}{2}\)

You are told that the original numerator was 2, so n=2 and \(\frac{2+5}{d+5}=\frac{1}{2}\). To find the original denominator all you need to do now is solve the equation. I'll leave that to you. I would start by multiplying both sides by d+5.

1.Re-writting this with fractional exponents gets: \(\frac{8^\frac{1}{4}}{4^\frac{1}{3}}\)

Now write 8 and 4 as powers of two, and then multiply the two exponents by the power-of-a-power exponent law (or whatever its called, been a while since I've done this) yields \(\frac{(2^3)^\frac{1}{4}}{(2^2)^\frac{1}{3}}=\frac{2^\frac{3}{4}}{2^\frac{2}{3}}\)

By the exponent division law (again, might have gotten the name wrong), you subtract the exponent in the denominator from the exponent in the numerator to get the exponent in the quotient.

\(\frac{2^\frac{3}{4}}{2^\frac{2}{3}}=2^{\frac{3}{4}-\frac{2}{3}}=\boxed{2^\frac{1}{12}}\)

http://artofproblemsolving.com/texer/

AoPS's TeXeR supports bbCode, LaTeX and Asymptote (what was used to create the drawing).

Asymptote is pretty cool, I've been able to make some fun stuff with it.

Triangle BPC is an isossilies triangle because sides BP and BC are the same length, so angles BPC and BCP are congruent. To find angle BPC you first need to find angle PBC.

Triangle APB is equilateral so all the angles measure 60 degrees, so angle APB measures 60 degrees. All the angles in a square measure 90 degrees. The measure of angle PBC is the sum of the angles ABC=90 and ABP=60 so PBC=150 degrees.

\(2 \cdot m\angle BPC+m\angle PBC=180\)

150 can be substituted for angle PBC

\(2 \cdot m\angle BPC+150=180\)

subtract 150

\(2 \cdot m\angle BPC=30\)

divide by 2

\(\boxed{m\angle BPC=15}\)

\( 3x^{\circ}+ 27^{\circ}, \space2y^{\circ} - 4^{\circ} \)

An equilateral triangle is a triangle where all the algles are the same, so they all measure 60 degrees.

The equations 3x+27=60 and 2y-4=60 can be made because they are the measures of the angles and both of them are equil to 60.

The equations can be solved to find x and y

3x+27=60

3x=33         subtract 27 from each side

x=11           divide by 3

2y-4=60

2y=64        add 4 to both sides

y=32          divide by 2

So x+y=11+32=43

The form of the equation of a circle is r²=(x-a)²+(y-b)², where r is the radius and (a,b) is the center.

The length of the line PQ is the diamiter of the circle, so plug (-1,2) and (7,8) into the distance formula

\(d=\sqrt{(-1-7)^2+(2-8)^2}\)

\(d=\sqrt{(-8)^2+(-6)^2}\)

\(d=\sqrt{64+36}\)

\(d=\sqrt{100}\)

\(d=10\)

so the radius is 5

The midpoint of the line PQ is the center of the circle, so plug P and Q into the midpoint formula

\(m=(\frac{-1+7}{2},\frac{8+2}{2})\)

\(m=(\frac{6}{2},\frac{10}{2})\)

m=(3,5)

Plugging the midpoint and the radius into the equation for the circle gets:

25=(x-3)²+(y-5)²

I don't see how this problem is flawed?

The numbers seem to work out.

I took the PSATthe end of last year. I found it to be easy (maybe because I had already taken the ACT), and had plenty of time to double and tripple check my work on the math section, and the reading and english sections also wern't that difficult. I atually just got my scores last week.

The PSAT is supposed to be just for practice, thats strange your school is making it affect your high school classes.

I checked out a practice book from the library, and I think just understanding the format of the test and the types of questions will help you immensly. Doing a timed practice test in each of the subjects will help you get the hang of managing your time and not spending too long on difficult problems.

If you are already a good student in math and english, and you know how the test works and can manage your time, you should be able to do well on the test with out doing that much to prepare.

Let m be the distance the squares overlap on the long edges.

First, express the area of the region the squares don't overlap in terms of the variable. (You can probabally skip this part, I'm just struggling to find more ways to use the variable :/ )

Area=2(7*(7-m))

The 2 comes from there being two rectangles, 7 is the height and 7-m is the width.

Now express the length of the total idth in terms of the variable.

14-m

With no overlap, the width would be 14, but to find the width after overlaping you have to subtract the distance they overlap

14-m=10

Set it equal to 10

-m=-4

Sutract 14 from each side

m=4

Divide by -1

Now that we have the distance they overlap, plug it into the equation from the beginning to get the answer.

2(7*(7-4))

2(7*3)

2*21

The area the squares don't overlap is 42

Yeah I want to take their classes too, although I go to a good math program in my area outside of school

I did take their AMC 8 weekend seminar a few years ago though