note that gradient = 
at x = a
calculate for each pair of functions and compare gradient
(a)
= 2x and = - 1
at x = 4 : gradient = 8 and - 1 : 8 > - 1
(b)
= 2x + 3 and = - 2
at x = 2 : gradient = 7 and - 2 and 7 > - 2
(c)
= 4x + 13 and = 2
at x = - 7 : gradient = - 15 and 2 and 2 > - 15
(d)
= 6x - 5 and = 2x - 2
at x = - 1 : gradient = - 11 and - 4 and - 4 > - 11
(e)
y = √x = 
= 1/(2√x) and = 2
at x = 9 : gradient = 
and 2 and 2 >
Answer:
y=2x-11
Step-by-step explanation:
Put it in the form y=mx+b=>y=2x-1/2. Instead of -1/2, put it as b, since you want it to go through 3,-5. y=2m+b when it goes through (3,-5) happens when -5=3*2-b. Do algebra, and get b as 11. That means the equation is y=2x-11.
Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.
![\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Ctextit%7Bhyperbolas%2C%20horizontal%20traverse%20axis%20%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B%28x-%20h%29%5E2%7D%7B%20a%5E2%7D-%5Ccfrac%7B%28y-%20k%29%5E2%7D%7B%20b%5E2%7D%3D1%20%5Cqquad%20%5Cbegin%7Bcases%7D%20center%5C%20%28%20h%2C%20k%29%5C%5C%20vertices%5C%20%28%20h%5Cpm%20a%2C%20k%29%5C%5C%20c%3D%5Ctextit%7Bdistance%20from%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bcenter%20to%20foci%7D%5C%5C%20%5Cqquad%20%5Csqrt%7B%20a%20%5E2%20%2B%20b%20%5E2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
Answer:
angle t is 55 degrees
Step-by-step explanation:
Here's my work:
Isoscelese means two of the same length sides. A triangle is 180 degrees. 180-70=110
110/2=55 (two because those are the two angles associated with the equal sides)
He still has 75% of his income left because 1500/375=4 so he spent 1/4 of his check and has 3/4 left.
