gb1falcon

This problem does not have a solution. To work, the answer must be a multiple of 4 as well as 1 more than a multiple of 6. These cannot both be true (one says even, the other says odd), so there is no solution.

Why are you asking this question a second time? You already asked it in the previous question. I answered it the first time, look there to see.

I will try to figure some of these out.

1. O is not given, neither is the center of the dilation.

2. A dilation divides all factors involved, including segment length, by its scale factor. So the answer is 20/-6, which you can simplify. A hint, though, whether a dilation is positive or negative only determines which side of the center of the dilation the result is on.

3a. 7 ways, since we can decide which point goes in any certain spot.

3b. 14 ways, since we can now reflect.

4. We can see that if A is 40 units away from B than it is 20 units away from the line. Given that since C goes to C it is on the line, how close can A be to C?

5. Most of the information does not matter since it does not change shape, only location. Use the dilation information I gave in problem two (just with scale factor -2 instead) to figure it out.

6. I don't know how to do this one.

7. If triangle ABC is equilateral, then the scale factor is -2. Otherwise, I don't think any dilation can do this.

The equation simplifies to \(44x^3+13x^2-24x-48=0\).

The solution for that is about 1, though not exactly.

See if you can figure it out yourself.

I'm not good at solving polynomials in the third degree.

1. Find all real numbers a that satisfy\(\frac{1}{64a^3 + 7} - 7 = 0\).

Move the seven to the right side.

\(\frac{1}{64a^3+7}=7\)

Multiply both sides by \(64a^3+7\).

\(1=448a^3+49\)

\(-48=448a^3\)

\(-3=28a^3\)

\(-\frac{3}{28}=a^3\)

\(a=-\sqrt[3]{\frac{3}{28}}\)

There is only one such real number.

2. Find all values of a that satisfy the equation\(\frac{a}{3} + 1 = \frac{a + 3}{a} + 1\).

Remove the 1:

\(\frac{a}{3}= \frac{a + 3}{a}\)

Multiply by 3a:

\(a^2=3a+9\)

\(a^2-3a-9=0\)

Solve using the quadratic formula:

\(\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:1\cdot \left(-9\right)}}{2\cdot \:1}\)

\(\frac{3\pm\sqrt{9-4\cdot-9}}{2}\)

\(\frac{3\pm\sqrt{9--36}}{2}\)

\(\frac{3\pm\sqrt{45}}{2}\)

\(\frac{3\pm3\sqrt{5}}{2}\)

You may choose what to do with that answer.

Simplify \(\sqrt[3]{576000 \cdot 4 \cdot 48}\).

Multiply the numbers:

\(576000\cdot4\cdot48=110592000\)

Cuberoot:

\(\sqrt[3]{110592000}=480\)

So 480 is the answer you are looking for.

I will assume that 'compounded continuously' means compounded annually. If this is not so, please tell me. Then

7500*1.029^14=11191.196 which rounds to $11190.

I will try a step-by-step solution. If the number is n, the equation is 

n-6=5(n+2)

n-6=5n+10

-6=4n+10

-16=4n

n=-4

Check:

-4-6=-10

5(-4+2)=-10

Therefore, -4 is the correct solution.