When you have something like this, all you need to do is substitute the values, the last is for what value of x
For the first one;
((x^2+1)+(x-2))(2)
(x^2+x-1)(2)
(2)^2+(2)-1
4+2-1
5
For the second one;
((x^2+1)-(x-2))(3)
(x^2-x+3)(3)
(3)^2-(3)+3
9-3+3
9
For the last one;
3(x^2+1)(7)+2(x-2)(3)
3((7)^2+7)+2((3)-2)
3(49+7)+2(3-2)
3(56)+2(1)
168+2
170
Answer
2p+12b=12
Step-by step explanation
I used p for plates and b for bowls\
2p+12b=4+(4*2)
2p+12b=12
I = / r where I = current and r = resistance
80 = k / 50 so
k = 400
so we have I = 400/r
when r = 40
I = 400/40 = 10 amps
Answer:
Step-by-step explanation:
equation of a circle
we know the center is (-7, 3)
the radius is 2
we replace (h,k) with the center coordinates
and input the radius
simplify to this:
