Check the picture below. So the parabola looks more or less like so.
let's recall that the vertex is half-way between the focus point and the directrix, at "p" units away from both.
Let's notice that the focus point is below the directrix, that means the parabola is vertical, namely the squared variable is the "x", and it also means that it's opening downwards as you see in the picture, namely that "p" is negative, in this case "p" is 1 unit, and thus is -1.
![\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \stackrel{\textit{we'll use this one}}{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-2\\ k=5\\ p=-1 \end{cases}\implies 4(-1)(y-5)=[x-(-2)]^2\implies -4(y-5)=(x+2)^2 \\\\\\ y-5=-\cfrac{1}{4}(x+2)^2\implies y=-\cfrac{1}{4}(x+2)^2+5](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bparabola%20vertex%20form%20with%20focus%20point%20distance%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%204p%28x-%20h%29%3D%28y-%20k%29%5E2%20%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7B4p%28y-%20k%29%3D%28x-%20h%29%5E2%7D%20%5Cend%7Barray%7D%20%5Cqquad%20%5Cbegin%7Barray%7D%7Bllll%7D%20vertex%5C%20%28%20h%2C%20k%29%5C%5C%5C%5C%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%20%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D-2%5C%5C%20k%3D5%5C%5C%20p%3D-1%20%5Cend%7Bcases%7D%5Cimplies%204%28-1%29%28y-5%29%3D%5Bx-%28-2%29%5D%5E2%5Cimplies%20-4%28y-5%29%3D%28x%2B2%29%5E2%20%5C%5C%5C%5C%5C%5C%20y-5%3D-%5Ccfrac%7B1%7D%7B4%7D%28x%2B2%29%5E2%5Cimplies%20y%3D-%5Ccfrac%7B1%7D%7B4%7D%28x%2B2%29%5E2%2B5)
<u>Given</u>:
Given that ABC is a right triangle.
The length of AB is 7 units.
The measure of ∠A is 65°
We need to determine the length of AC
<u>Length of AC:</u>
The length of AC can be determined using the trigonometric ratio.
Thus, we have;
Where the value of 
is 65° and the side adjacent to the angle is AC and the side hypotenuse to the angle is AB.
Substituting the values, we have;
Substituting AB = 7, we have;
Multiplying both sides by 7, we get;
Rounding off to the nearest hundredth, we get;
Thus, the length of AC is 2.96 units.
<h2>
Answer:</h2>
The graph is shown in the Figure below
<h2>
Step-by-step explanation:</h2>
In this exercise, we have an equation. On the left side we have a straight line with slope 
and there is no any y-intercept. On the right side, on the other had, we also have a straight line, but the slope here is 
. Therefore, by plotting these two straight lines, we have that the solution is the origin, that is, the point 
.
Answer:
The answer is -77
Step-by-step explanation:
Ok, so assuming by x2 you mean x squared, I will solve this. So basically when you have a function, f(-7) would mean that you would have to replace all the x's in the equation with -7. So let's write that out. that would be f(-7) = -7^2 + (-7*4). So now according to PEMDAS, you would solve the exponent first, and -7^2 is equal to -49, because when you solve it you would do -(7^2), which is -(49), which is then -49. So now you have f(-7)= -49+ (4*-7). Solving for (4*-7), you get -28. This leaves you with -49 + (-28), which is -49 - 28. Simplifying that, you get the answer, which is -77.
well and 300% increase means you multiply by 3 and then add the original number
in this case 300% of 25 is 75 and its an increase so 25+75=100
the answer is 100
