well, x-intercept means for a graphed funciton where the x-axis gets touched/intercepted, Check the picture below, when that occurs, y = 0, so if we want to know what "x" is at that point we can simply set y = 0 and solve for "x".
![-3x-5y=12\implies \stackrel{\textit{setting y = 0}}{-3x-5(0) =12}\implies -3x-0=12 \\\\\\ -3x=12\implies x=\cfrac{12}{-3}\implies x = -4 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{coordinates}}{(-4~~,~~0)}~\hfill](https://tex.z-dn.net/?f=-3x-5y%3D12%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bsetting%20y%20%3D%200%7D%7D%7B-3x-5%280%29%20%3D12%7D%5Cimplies%20-3x-0%3D12%20%5C%5C%5C%5C%5C%5C%20-3x%3D12%5Cimplies%20x%3D%5Ccfrac%7B12%7D%7B-3%7D%5Cimplies%20x%20%3D%20-4%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bcoordinates%7D%7D%7B%28-4~~%2C~~0%29%7D~%5Chfill)
Answer:
yeah he is right the answer is the photo
Step-by-step explanation:
Answer:
no no no! don't touch me there that is my no no square
Step-by-step explanation:
The answer to this question is:
A circle is growing so that the radius is increasing at the rate of 2cm/min. How fast is the area of the circle changing at the instant the radius is 10cm? Include units in your answer.?
✔️I assume here the linear scale is changing at the rato of 5cm/min
✔️dR/dt=5(cm/min) (R - is the radius.... yrs, of the circle (not the side)
✔️The rate of area change would be d(pi*R^2)/dt=2pi*R*dR/dt.
✔️At the instant when R=20cm,this rate would be,
✔️2pi*20*5(cm^2/min)=200pi (cm^2/min) or, almost, 628 (cm^2/min)
Hoped This Helped, <span>Cello10
Your Welcome :) </span>
