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substitute -1 for x and put it equal to -2

solve for p

then.

Check it is correct 

Use Pythagoras's theorem.

Show us your working and we will tell you if it is correct.

Hint:

Notice though that you are finding a little side so there will have to be a take away somewhere.

What have you done towards solving this problem?

Show us the working you have done so far.

Where did you get stuck?

What have you done so far to solve this by yourself?

A = 20 

B =   3        A – B  =  20 – 3  =  17 

C = 12  

D =   5        C + D  =  12 + 5  =  17  

_.

Please help me with this problem!!!

A fly is walking along the edges of a 3-inch cube.  If it starts at a corner, what is the longest it could walk without retracing an edge?

(x,y) = (4,-3).

Hi Guest!
 

This is just \(9+18+27+36+45+54+63=\boxed{252}\).

Or \(9(1+2+3+4+5+6+7)=9(28)=\boxed{252}.\)

I hope this helped you, guest!

:)

Hi Guest!
 

This is a right triangle, so the hypotenuse would be \(6a+1\) since the hypotenuse is the longest side.

So, \((a+1)^2+(6a)^2=(6a+1)^2\).

\(a^2+2a+1+36a^2=36a^2+12a+1\)

\(a^2+2a=12a\)

\(a^2=10a\)

So, \(a=0, \text{ or } a=10\)

But, a length cannot be 0, so \(\boxed{a=10}!\)

I hope this helped you, guest!
:)

The first term is 1 and each successive term is the sum of all the previous terms:

1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192

Although not true for the first term, each of the following terms has a value of 2^{n - 2}

where n is the number of the term.

We want       2^{n - 2}  >=  5000

              log(2^{n - 2})  >=  log(5000)

          (n - 2)log(2)  >=  log(5000)

                      n - 2  >=  log(5000)/log(2)

                      n - 2  >= 12.29

                            n  >=  14.29

So:  n  =  15

I believe that the number is 198.

Using these digits, two at a time (but there is no restriction that the same digit isn't used twice), the possibilities are:

11,18, 19, 81, 88, 89, 91, 98, 99

The sum is 594.

Dividing this by 3, gives us i98.

The ratio of peach area to the area of octagon is   1 : 2   

If  x = 3  then  | (3)² - 16 |  =  | 9 - 16 |  =  | -7 |  =  7

If  x = -3  then  | (-3)² - 16 |  =  | 9 - 16 |  =  | -7 |  =  7

-3 x 3  =  ...

If he never goes more than two days in a row without running, the minimum number of days that he runs is 4.

The schedule looks like this: NNYNNYN NYNNYNN

If he never runs more than two days in a row, the maximum number of days that he run is 10.

The schedule looks like this:  YYNYYNY YNYYNYY

[Notice that the schedules are the exact opposite.]

10 - 4  =  ????

Thank you!

If ABCD and EFGH are congruent, then EH = AD = 10.

Since the area of ABCD is 112, the area of EFGH is also 112.

The area of EFGH can be found by multiplying EH x EN.

Area(EFGH)  =  EH x EN

               112  =  10 x EN

              11.2  =  EN

You start with 64 houses and sell 1/4 of them ...

(1/4) x 64  =  16

So, 16 houses were sold.

This means that 64 - 16 = 48 houses are still left to be sold.

This month, 3/4 of these houses were sold.

(3/4) x 48  =  36

So, 36 houses were sold this month.

This means that 48 - 36 = 12 houses that remain to be sold.

Makes no sense

14/3·pi

Keep subtracting  2pi  from  (14/3)pi  until you have a value smaller than  2pi

(14/3)pi - 2pi  =  (8/3)pi

(8/3)pi - 2pi  =  (2/3)pi

The reason that you subtract  2pi  is that every  2pi  is one revolution around the circle.

Going around the circle for one whole revolution (or  2pi) results in the same values for sin, cos, and tan.

1. By stars and bars, the number of such polynomials is c(13,3) = 286.

2. By stars and bars, the number of solutions is c(26,4) = 14950.

You thought of the number 432.

64(0.5)ⁿ = 2

(0.5)ⁿ = 1/32

nlog_0.5(0.5) = log_0.5(1/32)

n = log_0.5(1/32) = 5

or you can just write it out:

64 --> 32 --> 16 --> 8 --> 4 --> 2

5 arrows = do it 5 times to get 2L