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g(x) = -2x^2 + 8x + 3.

x^2 + y^2 = 72.

The sum of the elements of S is 11.

20^18==2.62144e+23 = 2^36 * 5^18
Total number of prime factors= 54
Total number of ordered pairs= 2,701

20^18==2.62144e+23 = 2^36 * 5^18
Total number of prime factors= 54
Total number of ordered triplets= 4,104,919

(a^2)(b^3)=(20)^18.

20^18 =2^36 * 5^18

a =(2^18)^2 =262,144^2

b =(5^6)^3  =15,625^3

Actually, it does exist. 

The triple can also be a multiple of a primitive Pythagorean Theorem. 

For example, the triple \((8, 15, 17)\) has a perimeter of 40. We can scale this up by 4, to have the triple \((32, 60,68)\). 

The bases of this triangle are 32 and 60, so the area is: \(\Large {32 \times 60 \over 2} = \text{____}\)

A certain right triangle with integer side lengths has perimeter 160. What is its area?   

It's a right triangle,  Its sides are integers.  So the sides must be a Pythagorean Triple. 

No Pythagorean Triple totals 160 per https://en.wikipedia.org/wiki/Pythagorean_Triple

So this triangle doesn't exist. 

_.

We can write \({1 \over a^2} + {1 \over b^2}\) as \({a^2+b^2} \over (ab)^2\), where both a and b are the roots of the equation. 

Using Vieta's, we find the denominator is \({4 \over 5} ^2 = {16 \over 25}\)

Using the quadratic formula, we find that the 2 roots are: \({ 11 \pm \sqrt{41} \over 10}\). 

Plugging it in, we get this mess: \(\huge{{1 \over { 11 + \sqrt{41} \over 10 }^2} + {1 \over { 11 - \sqrt{41} \over 10 }^2}} \over \huge{ 16 \over 25}\)

Simplifying the numerator, we get \(\large{{33 \over 10} \over {16 \over 25} }\). This simplifies to \(\color{brown}\boxed{ 81 \over 16}\)

I get 1/28

Use the formula A = (a)(b)/2 to find its area, but you haven't given me the sides of the triangle and only the perimeter so I can only guess what the length of each side is.

Because they are on the same bases, we can disregard them. 

Now, we have the equation: \(AAA + BBB + 111 = 666\)

This means \(A + B = 5\)

The only possible cases are \((1,4)\), \((2,3)\), \((3,2)\), and \((4,1)\). Thus, are only \(\color{brown}\boxed{4}\) cases that 

Oh i see, good thinking! 

What is the longest possible value of AC?

Hello Guest!

\(\sqrt{12^2-x^2}+\sqrt{13^2-x^2}=13\\ x=\dfrac{12\sqrt{133}}{13}\)

 \(The\ longest\ possible\ value\ of\ AC\ is\ {\color{blue}2x}=2\cdot \dfrac{12\sqrt{133}}{13}=\color{blue}21.291\)

  !

It would be sensible to write them in order from smallest to largest then you will not have to think about the final comparisons so much.

The bigger the power of 10 is then bigger the number is.

             Assuming it is multipled by a number between one and 10 (scientific noatation)

So looking at the first one

2.1x10^-4     

2.1 is between 1 and 10 so  this number is in scientific noation.  (All of them are in scientfici notation.)

Any number in scientific noation with a power of 10 less than -5 will be less than that one.

Si it is definitely noNOT Less than 8.2x10^-5

Is it Greater than 8.2x10^-4    Well they hae the same power, but 8 .2 is bigger than 2.1 so NO

Is 2.1x10^-4   Between 8.2x10^-4 and 8.2x10^-5 ?

Why do't I put the little one first ...    Is it Between 8.2x10^-5  and 8.2x10^-4 

Its bigger than the little one and smaller than the big one so it has yo be between them!

You can do the others yourself.

If you want to discuss you answers/working that is fine.

Using the distance formula, we have the following system, where:

 \(x = \) x co-ordinate of the point

\(y = \) distance between 2 points

\(64+x^2=y^2\)  Distance between \((0,8)\) and \((x, 0)\)

\(y=x+2\)        Distance between \((-2,0)\) and \((x,0)\)

Substituting in the second equation, we have: \(64+x^2=x^2+4x+4\)

Solving for x, we find \(\color{brown}\boxed{x = 15}\)

What was the total cost?

Hello Guest!

\(\$967\cdot 95\%\cdot 101\%=\color{blue}\$927.84\)

Enter into the calculator with \(\boxed{\%}\).

  !

Since the median is ten and there are ten distinct numbers, the median must be found by finding the mean between the two "middle" numbers.

This could be done by having the integers 9 and 11 in the group.

The four smallest numbers less than 9 are 1, 2, 3, and 4.

The four smallest numbers more than 11 are 12, 13, 14, and 15.

My guess is that the set consists of 1, 2, 3, 4 9, 11, 12, 13, 14, and 15.