All Answers 358502 Answers

\(9^y = {3*6*27*4\over81*2}\)

\(9^y = {81 * 6 * 2 * 2\over81 * 2}\), cancel terms (81 and 2)

\(9^y = {1 * 6 * 2 * 1\over1*1}\)

\(9^y = 12\)

y is approximately 1.131

Combine like terms:

x^2 + y^2 + 20x - 8y - 48 = 0

(x + 10)^2 + (y - 4)^2 - 164 = 0

radius is square root of 164.

Area = \(2\pi\sqrt{41}\)

The area is 24*pi

[DEF]/[ABC] = 5/11

You can combine these two, since the common ratio is the same.

6, a, b, 1296.

1296 = 6 x r^3

216 = r^3

r = 6

a = 6x6 = 36

This needs to be minimized: \({13\over 1-r}\).

Where -1 < r < 1. (If that isn't true then the series doesn't really have a sum)

If r is -1, the sum is 13/2 = 6.5. Rounding this figure up will both give an integer and put r back into its accepted range, so the smallest integer that can possibly be the sum of an infinite geometric series whose first term is 13 is 7. (the common ratio would be -6/7)

\({d+1\over d} \times {d+2\over d+1} ... \times {d+20\over d+19}=3\)

The numerator of the first term cancels the denominator of the second, and so on until the numerator of the 19th term cancels the denominator of the 20th.

You are left with d+20/d =3

3d = d + 20

2d = 20

d = 10

The original fraction is d+1/d = 11/10

If the parabola opens upwards, then a must be positive.

The roots will always be 3 and -2. This gives a base of 5.

5h/2 = 12

h = 24/5

However, since the parabola opens upwards, the intersection with the y-axis will be negative.

The graph of the parabola must have a y intersect at -24/5. Plugging this point in:

-24/5=a(-3)(2)

-24/5=-6a

4/5 = a

This is true: \(a_n^2 - a_{n + 1} = (-1)^n\)

And we have \(a_1 = 5\)

So this must be true: \(5^2-a_2=(-1)^1\), or \(26 = a_2\).

Similar logic:

\(26^2-a_3=(-1)^2\)

\(\fbox{$a_3=675$}\)

x * (100-20)% = 500

x * 80% = 500

4/5 * x = 500

4x = 2500

x = 625

(All of these values are found using simple area formulas, or length formulas derived from the Pythagorean Theorem)

AB is the square root of 2.

The area of the semicircle is pi/2.

The area of the square is 2.

The area of the circle around the square is pi.

The area of the part of the circle between arc AB and AB is (pi-2)/4

The area of the lune (shaded part) is pi/2 - (pi-2)/4 = pi/4 + 1/2

(I think)

The sidelength of the square inscribed in ABC has to be half of BC.

The sidelength of the square inscribed in DEF has to be a third of DF.

The ratio of BC:DF is 1:square root of 2

The ratio of 1/2:1/3 is 3:2

The ratio of the sidelength of the square inscribed in ABC to that of the square inscribed in DEF is 3:2 times the square root of 2.

Therefore, the ratio of their areas is 9:8.

 15 x 8/9 = 40/3

f(2) = -2. 

I don't think this requires explanation.

The vertex will come three units to the left compared to the original.

(-1, -1)

Ok so first, what I did was simplify the denominators of the fractions. 3^2-1 is 8, 5^2-1 is 24, and 7^2-1 is 48. (Because 3 squared is 9, 5 squared is 25, and 7 squared is 49.)

So the fractions are 6/8, 6/24, and 6/48. Then I simplified them. To make it easier, I simplified them into eighths.

6/24 = 2/8 (We could've put 1/4, but I'm gonna use eighths for this problem. You can use whatever you want though.)

6/48 = 1/8

After that we add 6/8 + 2/8 + 1/8 to get 9/8.

So the answer is 9/8, or 1 1/8.

f(2) = 6

The shaded area is (pi - 1)/2.

Yes, LPQ is the triangle we want the area of. I might be able to post a picture of the question, if you'd like. 

(a) ab = 24

(b) a + b = 11

(c) (a,b) = (3,8) and (8,3)

The answer is -13.