Improve Your Skills with Our Coordinate Plane Practice Problems

I am stuck here on this problem, it should be in the six stepts of quadratic equations.
$$\begin{gathered} a{x}^2-<!--\; -\; -->2x-<!--\; -\; -->13=0 \\ {x}^2+2x=13 \\ 4x^2+8x=52 \\ 4x^2+8x+4=52+4 \\ 4x^2+8x+4=56 \end{gathered}$$

Determine: $x^2-<!--\; -\; -->9x+14$

How old am I if 400 reduced by 2 times my age is 244?

Show that the points form the vertices of the indicated polygon.
Right triangle: (-1,6),(3,8),(5,4)

Find the measure of angle whose supplement is eleven times its complement.
Hint: Supplement and complement of an angle are $(180-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}{)}^{\circ <!--\; \circ \; -->}\text{and}(90-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}{)}^{\circ <!--\; \circ \; -->}$ respectively

The mathematician christian Goldbach (1690-1764) conjectured that every odd number greater than 5 is the sum of three primes. verify this conjecture for the follwing numbers: 25, 33, 49

Use all four of the numbers and any operation. Use 2,3,4,6 to equal these numbers:
41, 47 and 53
Rules:
1. It must follow the order of operations.
2. You can use any operation: $+,-<!--\; -\; -->,\times <!--\; \times \; -->,÷<!--\; ÷\; -->$
3. You can also use grouping symbols (parentheses)
4. You can also use ANY operation on a calculator.
5. Exponents, factorials, decimal points, and square root are all allowed.
6. The Tricky Part, you have to use ALL 4 numbers and can not use a number more than once.
7. Remember, they are digits, so 3 and 6 can not be made INTO the number 36.
8. Reminder, using the number two as an exponent COUNTS as you using the number two.

I know this may be silly but lately I've been wondering about powers of negative numbers and I came up with an ambiguous example:
$(-<!--\; -\; -->1{)}^{(\frac{22}{10})}$
I wanted to know if it was +1 or -1. My first attempt was writing it down as a root:
$\sqrt[10]{(-<!--\; -\; -->1{)}^{22}}$
Minus one to the even power is obviously plus one, then 1 to the power of $\frac{1}{10}$ is 1. Yet if we first reduce the fraction and then calculate we get a different answer:
$(-<!--\; -\; -->1{)}^{(\frac{22}{10})}=(-<!--\; -\; -->1{)}^{(\frac{11}{5})}=\sqrt[5]{(-<!--\; -\; -->1{)}^{11}}$
Even though it's (I suppose) the same situation, $(-<!--\; -\; -->1{)}^{11}$ is and we end up with $\sqrt[5]{-<!--\; -\; -->1}$, a complex number. What is (or is there even) the value of $(-<!--\; -\; -->1{)}^{(\frac{22}{10})}$?

Is there a number whose absolute value is negative?
I've recently started to think about this, and I'm sure a couple of you out there have, too.
In Algebra, we learned that $|x|\ge <!--\; \ge \; -->0$ , no matter what number you plug in for x. For example:
$|-<!--\; -\; -->5|=5\ge <!--\; \ge \; -->0$
We also learned that $x^2\ge <!--\; \ge \; -->0$. For example:
$(-<!--\; -\; -->5{)}^2=25\ge <!--\; \ge \; -->0$
The exception for the $x^2$ rule is imaginary numbers (which we learn later on in Algebra II). Imaginary numbers are unique, in that their square is a negative number. For example:
$4i^2=-<!--\; -\; -->4$
These imaginary numbers can be used when finding the "missing" roots of a polynomial equation.
My question to you is this: Is there any number whose absolute value is negative, and how could it be used?

When solving inequality equations, we have to change the sign of the inequality when we divide/multiply by a negative number. And I get why. Because your making one side positive and one side negative.
But let's say you were multiplying by say $(x^3+1)/x>1$ or something. When you multiply both sides by x, why don't you need to flip the sign? Because there's the possibility x is negative! Why don't you need to construct to equations, one where the sign is the same, and one where the sign is flipped.
As an extension, let's say you were multiplying both sides of $(x^3+1)/-<!--\; -\; -->x>1$ by -x. Do you need to change the sign too?

Why does finding the union of these three sets yield a negative number?
I've been working on a homework problem that I can't seem to be able to solve.
The question states:
Suppose 25 people attended a conference that contains 3 sessions. 15 people attended session #1; 18 people attended session #2; 12 people attended session #3. At least how many people attended all 3?
$u,\sin <!--\; \; -->2u=-<!--\; -\; -->2\sin <!--\; \; -->u\cos <!--\; \; -->u$
However, when I'm applying this problem to 3 sets, I obtained a negative number.
I computed the result: |A∪B∪C| ≤ 5 where |A∩B| ≥ 8, |A∩C| ≥ 2, |B∩C| ≥ 5, however, I don't quite understand the negative part where 25 ≥ |(A∩B)∪C| = |A∩B| + |C| - |A∩B∩C| (shorter way to find the minimum). The result just ends up as |A∩B∩C| + 25 ≥ 20 which would result in a negative unless I can somehow divide both sides by -1 which would change the equality?

Why is 7 handled as a negative number in the following equation?
Here is a very simple multi-step equation that bothered me for a specific reason:
20-7x=6x-6
I started by subtracting 6x from both sides and for a second I thought I would find "1x" as a result of subtracting 6x from the left side of the equation (and write it as 20-1x=-6). then I realized that it should have been "-13 x" (20-13x=-6)
Although I could solve it mechanically, I couldn't internalize the logic behind evaluating a subtraction sign as a negative number symbol in algebraic equations. Why should one get the "-" sign before 7x as a negative sign and conduct the subtraction operation accordingly? I would appreciate an answer that covers the logic behind this action.

I was given a problem to evaluate the following integral,
$\underset{-<!--\; -\; -->4}{\overset{-<!--\; -\; -->1}{\int <!--\; \int \; -->}}{\displaystyle \frac{1}{x}}dx$
I know the fact that integration of $1/x$ is $\log <!--\; \; -->(x)$ but how can we compute negative logarithm? I mean to say that since definite integral refers to the area under the curve so how can it be even defined?
I tried the following method:
$\underset{-<!--\; -\; -->4}{\overset{-<!--\; -\; -->1}{\int <!--\; \int \; -->}}{\displaystyle \frac{1}{x}}dx=\log <!--\; \; -->(-<!--\; -\; -->1)-<!--\; -\; -->\log <!--\; \; -->(-<!--\; -\; -->4)$
$⟹<!--\; ⟹\; -->\log <!--\; \; -->\left({\displaystyle \frac{-<!--\; -\; -->1}{-<!--\; -\; -->4}}\right)=-<!--\; -\; -->\log <!--\; \; -->(4)$
Is it correct? I don't think that the answer should be $-<!--\; -\; -->\log <!--\; \; -->(4)$ because there is no graph for the negative logarithm. Is there any need of complex logarithm?

the sine function is an odd function, for a negative number $u,\sin <!--\; \; -->2u=-<!--\; -\; -->2\sin <!--\; \; -->u\cos <!--\; \; -->u$. Is it true or false and why?

Trivial ODE and trivial question:
$x(y+4)+\frac{dy}{dx}=0$
with initial conditions $y=-<!--\; -\; -->5,x=0$
After we separate variables we get:
$-<!--\; -\; -->\frac{dy}{y+4}=xdx$
Integrate left and write parts:
$-<!--\; -\; -->\ln <!--\; \; -->(y+4)=\frac{1}{2}{x}^2+C$
Here we see that if $y=-<!--\; -\; -->5$ we have a log of negative number. Trying not to think about it, I proceed as follows:
Exponentiation of both parts:
$y=e^{-<!--\; -\; -->1/2x^2-<!--\; -\; -->C}-<!--\; -\; -->4$
And then:
$y=e^{-<!--\; -\; -->1/2x^2}{e}^{-<!--\; -\; -->C}-<!--\; -\; -->4$
$y=e^{-<!--\; -\; -->1/2x^2}C-<!--\; -\; -->4$
(this new $C$ to $e^{-<!--\; -\; -->C}$ and cannot be negative)
$-<!--\; -\; -->5=e^{0}C-<!--\; -\; -->4$
$C=-<!--\; -\; -->1$
(but we see that it is in fact negative under given initial conditions)
which gives us the correct answer:
$y=-<!--\; -\; -->e^{-<!--\; -\; -->1/2x^2}-<!--\; -\; -->4$
Now I feel that I've cheated somewhere. My only guess that I was able to deal with a log of a negative number and get negative C (which shouldn't be negative) is because I have involved complex numbers between the lines. Is it so? If not, how can I make this solution rigorous?

let $n$ is non-negative number so the equations$(x^2+1{)}^2+n=yz+1$$(y^2+1{)}^2+n=zx+1$$(z^2+1{)}^2+n=xy+1$have $(x,y,z)$ real solution.find all solutions for the non-negative n that make $(x,y,z)$ are real numbers and find $(x,y,z)$ also

time (min)0, 2.2, 3.4, 4,4. 6 height (min) 14, 80, 116, 134, 152 A.write a linear function to model the situation (show your solutions)

Question
Write the product using exponents.
(-t) X (-t) X (-t) X (-t) X (-t)

Where would the equations 2x + 15y = 13 and -3x + 5y = 8 intersect on a coordinate plane?

Natural log of a negative number
${\displaystyle \ln (-1)=\ln \left(e^{i\pi }\right)=i\pi }$