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\(\text{Consider a line from the right bottom corner to the center of the circle}\\ \text{Convince yourself that it bisects the angle the triangle makes with the right side}\\ \text{and thus it is a }15^\circ \text{ angle}\)

\(\tan(15^\circ) = \dfrac{r}{s-r}\\ s = \dfrac{r(1+\tan(15^\circ))}{\tan(15^\circ)}\)

\(\text{I'll let you plug and chug to see that it's }s=18\)

\(f(x)=4x^2+7x+k \\ \text{has roots }x = \dfrac{-7\pm i\sqrt{15}}{8}\\ b^2-4ac=-15\\ 7^2 - 4(4)k = -15\\ -16k=-64\\ k=4\)

Ok, I spelled eighteen wrong.

9 isn't in the range either.  So 14 are in the domain and not in the range.

Sorry for the confusion.

\(\text{What's described can happen in the following ways}\\ \text{All 3 tiles match on a letter}\\ \text{All 3 tiles match on a number}\\ \text{1 of the tiles match both the other two, but the other two do not match}\\ \text{There are three ways that the previous line can occur}\\ [(1,2),(1,3)],[(1,2),(2,3)],[(1,3),(2,3)]\)

\(P[\text{3 tiles match letter}]=\dfrac{4}{24}\dfrac{3}{23}= \dfrac 1 2 \dfrac{1}{23}\\ P[\text{3 tiles match number}] = \dfrac 1 2 \dfrac{1}{23}\\ P[\text{2 tiles match the other tile but not each other}] = \dfrac{8}{24}\dfrac{4}{23} = \dfrac 4 3 \dfrac{1}{23}\\ P[\text{what's described}] = 2 \times \dfrac 1 2 \dfrac{1}{23} + 3 \times \dfrac 4 3 \dfrac{1}{23} = \dfrac{1}{23}+\dfrac{4}{23} = \dfrac{5}{23}\)

1)

\(a\cdot g(x) \text{ will always have degree 4 for }a\neq 0\\ \text{assuming no cancellation the sum will have degree 4 as well}\\ \text{this is the largest degree the sum can have}\)

2)

\(\text{let }b=-\dfrac 1 2\\ f(x)+b\cdot g(x) = (1-1)x^4 + (-3+3)x^2 -x+\left(2+\dfrac 1 2\right)=\\ -x+\dfrac 5 2 \text{ which has degree 1}\)

Hmmm...That was incorrect...

\(r=0.1\\ n_1=4,~n_2=1\\ p=10000\\ \text{plan 1}:\\ payment_{1,1} =\dfrac 1 2 p\left(1+\dfrac{r}{n_1}\right)^{5n_1}\\ payment_{1,2} = payment_{1,1}\left(1+\dfrac{r}{n_1}\right)^{5n_1}\\ p_1 = \text{Plan 1 total payment = }payment_{1,1} + payment_{1,2}\)

\(\text{Plan 2}:\\ payment_{2,1} = p\left(1+\dfrac{r}{n_2}\right)^{10 n_2}\\ p_2=\text{Plan 2 total payment = }payment_{2,1}\\ \text{find }|p_2-p_1|\)

Hint: notice the equation is in standard form. Add the -4 to the other side

-4x+8y=4

Now use your knowledge on standard form or convert into slope-intercept form (y=mx+b)

\(r_{row}=\dfrac{1rm}{14hr}\\ r_{rub}=\dfrac{1rm}{6hr}\\ x \cdot r_{row} + y \cdot r_{rub} = \dfrac 1 2\\ y \cdot r_{row} + x \cdot r_{rub} = 1\\ \dfrac{x}{14} + \dfrac{y}{6} = \dfrac 1 2\\ \dfrac{y}{14} + \dfrac{x}{6} = 1\)

Do you think you can solve it from here?

There's no comprehensive formula for the number of letters in the English spelling of a given number.

So you've got to just list them all out and check.

\(\text{The range of }\ell(n) \text{ is }(3,4,\dots,8)\\ \text{Thus }|(1,2) \cup (9,10,\dots, 20)| = 14\\ \text{is the number you are after}\)

\(z = r e^{i \theta}\\ \sqrt{z} = \sqrt{r} e^{i \theta/2}\\ -i = 1\cdot e^{-i \pi /2}\\ \sqrt{-i} = \sqrt{1} e^{-i \pi/4} = e^{-i\pi/4} = \\ \cos(-\pi/4) + i \sin(-\pi/4) = \\ \dfrac{\sqrt{2}}{2}(1 - i)\)

This is just base angle theorem. (if the base angles are equal, the triangle is isosceles, and by definition QS = QR)

Let the boxes be Box 1, Box 2, Box 3. 

consider the 3 white balls. They can be all of them in one box:

(3, 0, 0) (3 in Box 1, 0 in box 2 and 0 in box 0)
(0, 3, 0)
(0, 0, 3)

We can have 2 in one box, and 1 in one of the remaining boxes:

(2, 0, 1)
(2, 1, 0)
(0, 2, 1)
(1, 2, 0)
(0, 1, 2)
(1, 0, 2)

and there is only one way: (1, 1, 1) to place one white ball in each box

In total there are: 3+6+1=10 ways to place the white balls. Similarly there are 10 ways to place the black ones.

Since every placement of the white balls can be combined with any placement of the black balls, there are 10*10=100 ways to place the 3 white balls and the 3 black bals in the boxes.

Or if you just want the answer, its 100

Thank you, I understand now.  :)

Yes, As I read it, x/2 is for me x divided by 2. As x/2 = 8, x divided by 2 is 8, and 16 divided by 2 is 8, so X=16

See last answer.

hint for both.

draw a right anglesd triangle.

Label the right angle, and call the other angles theta and beta.

Label the hypotenuse h

call the other 2 sides p and q   doesn't matter which is which.

Now what is      cos(theta)   

and what is       sin(beta)

What is    theta+alpha ?

What do those word in the answers mean?  You can look them up online by yourself if you need to.

Please no one answer over me.

Look at the working for the last answer you were given.

Learn from it.  

What do you think that the answer might be?

Please no one answer over me.

The graph is 3 units higher than it was before so what do you think the anwer might be?

Please no one answer over me.

What do you think the answer is?

Talk about your question and then we can help.

Please no one answer over the top of me.