Answer:
p>-1 or p<-19
Step-by-step explanation:
4p-10<6p-8 (add 10 to each side)
4p< 6p+2 (subtract 6p from each side)
-2p<2 (divide by -2 and switch the sign)
p>-1
10p+16<9p-3 (subtract 16 from each side)
10p<9p+-19 (subtract 9p from each side)
p<-19
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)

Hey there!
They are both correct, and their expressions are equivalent.'
9(S + T) + 45 = 9S + 9T + 45
Have a terrificly amazing day!
May is the 10th month after july
y=−3x+9 slope of this equation is -3 and y-intercept is 9
Plot the y -intercept first which is 9 and mark the points using the slope which is -3. Join the points. So blue line in the graph is y = -3x+9
y=−x−5 slope of this equation is -1 and y intercept is -5
Plot the y -intercept first which is -5 and mark the points using the slope which is -1 . Join the points. So pink line in the graph is y = -x-5
The intersection point of these two lines gives you the solution.
The coordinates of point of intersection (7,-12)