WhyamIdoingthis

Let the first number be x and the second one be y.

\(x+y\over2\)=75

\(\sqrt (xy)\)=21

(yes the xy is under the radical)

The equations can be manipulated into:

xy=441

x+y=150

From here you would have two options: graph, or solve using quadratics.

Let x be z².

The equation turns into this:

x²-4x+4=0

This factors as:

(x-2)²=0

x would be equal to two. Set 2 equal to z² and solve:

\(z=\sqrt2 \) \(or \)\(-\sqrt2\)

If p and the square of q are inversely proportional, p x q x q = some constant k. When p is 19 and q is 8, k is 1216. k remains the same throughout all pairs of (p,q). If q is 3,

p x 9 =1216

So p would be equal to \(135 \)\(1 \over 9\).

Solution: \(1\)

Answer:

We are looking for the number of sevens there are in the prime factorization of the number.

60! has 9 sevens in its prime factorization, because there are eight multiples of seven in 1x2x3x4...x60, and one of them is 7².

Because the 60! is being square, the number of sevens in its prime factorization doubles to 18.

120! has 19 sevens in its prime factorizations, because there are 17 multiples of seven in 1x2x3x4....x120, and two of them are multiples of 49.

\(x \cdot7^{18}\over y \cdot 7^{19}\)\(=x/y \cdot 7\)

where x and y are two numbers that aren't multiples of 7 (and so therefore neither is their quotient).

There is only one seven the prime factorization of this number, which means that the largest integer n such that 7^n that divides it is just \(1\).

Answer: \(x=-7\), \(y=-2.8\)

Solution:

Adding both equations gives -x=7, so x=-7.

Plugging that back into the first equation gives -5y=14, so y=-2.8.

Answer: 10

Explanation:

Simplifying that expression turns it into \(x^2+(k-5)x+36\). The 36 at the end (and the fact that the coefficent of x^2 is 1) shows us that there are only two possible options: (x+6)^2 and (x-6)^2. Expanding the first gives x^2 + 12x +36, while the second gives x^2-12x+36.

Set up an equation for the first one:

k-5=12

k=17

Set up an equation for the second one:

k-5=-12

k=-7

-7 + 17 = 10.

There are really only two numbers that satisfy this: 124 and 248. (The next possible one would be 36{12} which doesn't work in a base ten system.)

248 is the larger one and is therefore your answer.

Answer: 5

Explanation:

(EDIT: The slow way)

I'm going to immediately say that however I solve this is going to be a complete waste of time and that there is almost certainly a smarter way to do this that I can't think of at the moment. But at least I'll get an answer.

Start by expanding \(2(x-5)^2-8\) into the given form:

\(2(x-5)^2-8\)

\(=2(x^2-10x+25)-8\)

\(=2x^2-20x+42\)

It should be clear now that a=2, b=-20, and c=42. So the new expression with the tripled values looks something like this:

\(6x^2-60x+126\)

Now all that needs to be done is converting this into \(n(x-h)^2+k\) form. Already we can tell that n=6.

\(6(x-h)^2+k=6x^2-60x+126\)

\(6x^2-12hx+h^2+k=6x^2-60x+126\)

There is only one term in each side which contains an x^1, namely -12hx and -60x. They must equal each other, and so \(h=5\).

The quicker way:

h didn't change from the two equations, suggesting that there was a quicker way to solve this question.

In this particular question, the x^2 coefficient becomes tripled, the x coefficient becomes tripled, and the integer value becomes tripled.

k is irrelevant to h, so we leave it out of this. Only the \(n(x-h)^2\) part matters to h.

There is one variable present in all three terms of the expanded version of the expression, namely n. So the only number needed to be changed in this part of the form is n. h doesn't change, so if the original value of h was five, the new value is also 5.

I have no idea why I didn't see that before.

Answer:\(-102\)

Explanation:

The slope of the given line is -2/8, which reduces to -1/4.

Using the slope formula between the two given points, we get an equation:

\(\frac{17-(-9)}{j-2}\)\(=-\frac{1}{4}\)

\(-104=j-2\)

\(j=-102\)