geno3141

The problem (x⁵y²) / (yx²) can be written as:  (x·x·x·x·x·y·y) / (y·x·x)

and can be simplified by cancelling out the same factors in the numerator and denominator to end with:  x³y

(Or, since you are dividing, subtract the exponents from the same factors.)

If the three sides have lengths 7, 8, and 15, they will not form a triangle; they will lie on one line segment.

For three lengths to form a triangle, each two lengths added together must be larger than the third length, in all possible combinations.

Since there are 24 hours in a day, multiply the number of days in the month by 24.

2^{2x + 13}  =  3^{x - 33}

Take the log of both sides:

log(2^{2x + 13}) =  log(3^{x - 33})

Since exponents come out as multipliers:

(2x + 13)log(2)  =  (x - 33)log(3)

Use the distributive property:

2xlog(2) + 13log(2)  =  xlog(3) - 33log(3)

Rearrange:

2xlog(2) - xlog(3)  =  -13log(2) - 33log(3)

Factor out the x:

x( 2log(2) - log(3) ) =  -13log(2) - 33log(3)

Divide:

x  =  [ -13log(2) - 33log(3) ] / [ 2log(2) - log(3) ]

1) Go to a teacher that you trust.

2) Go to a guidance counselor.

3) Go to an assistant principal, or the principal.

4) Write down what the bullying teacher says and does and who is present. 

You can interpret "90% of the original price is $24.50" into algebra as:  "0.90 · x = 24.50".

"90%" becomes "0.90"

"of" becomes "times" becomes "·"

"is" becomes "="

Just given one angle measurement -- and nothing else -- you do not have enough information to find area. However, in a triangle, if you know the size of one angle and the sizes of the two sides that include that angle, you can find the area of that triangle by using the formula;  A  =  ½·a·b·sin(C).

'a' and 'b' are the lengths of the two sides that include angle C.

The volume of a rectangular pyramid is (1/3) x (Area of the base) x (height)

When depressed, hang in there, for the probability of getting better is much higher than the probability of getting worse -- and, when it gets better, you'll be happy that you held on ...

I think that you'll have to try that question again ...