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Im sorry, What!!??

why?

Qeue

With a .

if u mean suicide then dont. that is never the answer

yes

1x1x5= 5

Ill go im only 13 though so you will be safe and not be lonley but if youve been asked 5 times go with the nerdiest one make him feel wanted

1+1=2
Proof:

Easy

The Pythagorean Theorem, which is the square root of the sum of 2 side measures squared. \(\sqrt{a^2+b^2}\)

Fill in the values of the sides and solve to find the side C, or the hypotenuse. Make sure to use a calculator when you square root, because even if it is a long decimal that is the better answer.

By "oval," I'm assuming that you might mean "ellipse"

The area of an ellipse is given by  pi* a * b    where a and b are the lengths of the semi-major and semi-minor axis

Since the ovals are similar in all respects and the diameter of the larger one to the diameter of the smaller one is 22.5/7.5  = 3 / 1  ....then the dimensions of the larger are 3 times as much as the smaller

Then....the area will be 9 times as great because the lengths of both the semi-major and semi-minor axis will be 3 times as great....so the area of the larger =  9 * 11.7  = 105.3 cm^2

Mistake  9711.167542

971.1167541

Use the FOIL Method.

x*x=x²

x*(-2)=-2x

x*2=2x

2*(-2)=-4

x²-2x+2x-4

Now combine like terms

-2x+2x=0x. 

Since the middle of the trinomial is 0x, there's no need to write it in.

x²-4

The answer is x²-4

There really is no such number......here's why.....

If the number consisted of only one prime factor....it would have to be either a square, a cube, or 3^4 = 81....

But.....the only cube  that exists between 50 and 100  is 4^3 = 64  = 2^6

And  the divisors of 64  are  1 | 2 | 4 | 8 | 16 | 32 | 64 ......however, we can only form 4 possible rectangles with these divisors -  1 x 64, 2 x 32, 4 x 16 or 8x 8

So......the number must be a square......but the only squares between 50 and 100 are 64 and 81 and....we have seen that 64 isn't possible

And the divsors of 81 are 1 | 3 | 9 | 27 | 81  .........and only 3  rectangles are possible......1 x 81,  3 x 27 and 9 x 9...... [note, for the same reason, 3^4 = 81   isn't possible, either ]

80 would provdie us with 5 rectangles because its dvisors are 1 | 2 | 4 | 5 | 8 | 10 | 16 | 20 | 40 | 80  

And the rectangles are : 1 x 80, 2 x 40, 4, 20, 5 x 16 and 8 x 10

But 80 has two prime factors [ 2 and 5]  because 80  =  2^4 x 5

So.....no number exactly matches the given criteria......

the answer for x is 5 

10.10=100

100

John and Nat were given some money.
If John spends $50 and Nat spends $100 each day,
John would still have $2500 left while Nat would have spent all her money.
If John spends $100 and Nat spends $50 each day,
John would still have $1000 left while Nat would have spent all her money.

How much were John and Nat given each?

Let j = Johns money at start

Let n = Nats money at start

Let x = days until Nat spent all her money first run.

Let y = days until Nat spent all her money second run.

1. Run

\(\begin{array}{|rcll|} \hline j - x\cdot 50 &=& 2500 \\ j &=& 2500+50x\\\\ n - x\cdot 100 &=& 0 \\ n &=& 100\cdot x \\ \hline \end{array}\)

2. Run

\(\begin{array}{|rcll|} \hline j - y\cdot 100 &=& 1000 \\ j &=& 1000+100y \\\\ n - y\cdot 50 &=& 0 \\ n &=& 50y \\ \hline \end{array}\)

We set equal:

\(\begin{array}{|rcll|} \hline n = 50y &=& 100x \\ y &=& 2x \\\\ 1000+100y &=& 2500+50x \quad & | \quad -1000\\ 100y &=& 1500+50x \\ 100\cdot(2x) &=& 1500+50x \\ 200x &=& 1500+50x \quad & | \quad -50x\\ 150x &=& 1500 \quad & | \quad :150\\ x &=& \frac{1500}{150} \\ x &=& 10 \\\\ n &=& 100x \\ n &=& 100\cdot 10 \\ \mathbf{ n } & \mathbf{=} & \mathbf{$ 1000 }\\\\ j &=& 2500+50x \quad & | \quad x = 10\\ j &=& 2500+50\cdot 10 \\ j &=& 2500+500 \\ j &=& 3000 \\ \mathbf{ j } & \mathbf{=} & \mathbf{$ 3000 }\\\\ \hline \end{array}\)

John were given $3000 and Nat were given $1000

5+5=10

Because seven ate nine

At the time you start the thing, you are 1900-1879. Today you would be 2016-1879.

1.7724538509055159

25%  =   1/4

So...1/4 of the class likes strawberries......= 8 students......so......4 times this must be the total students in the class = 32