Answer:
Step-by-step explanation:
so you want to start by using y=mx+b
step one: find the slope (remember m is the slope in this)
m= 
m= 
m= 
m= - 
step two: insert the value we found for m into the equation and take one of the given points and insert the x and y into the equation (im using (0,5)) and solve for b.
y=mx+b
5= -4/3(0)+b
5=b
step three: rewrite y=mx+b with the values we found for m and for b
A quadratic function is given by y = ax^2 + bx + c
c(2) = a(2)^2 + 2b + c = 45
4a + 2b + c = 45 . . . (1)
c(4) = a(4)^2 + 4b + c = 143
16a + 4b + c = 143 . . . (2)
c(10) = a(10)^2 + 10b + c = 869
100a + 10b + c = 869 . . . (3)
Solving (1), (2) and (3) gives a = 9, b = -5, c = 19
Therefore, c(x) = 9x^2 - 5x + 19
c(7) = 9(7)^2 - 5(7) + 19 = 9(49) - 35 + 19 = 441 - 16 = 425
Therefore, it costs $425 to produce 7 calculators.
[ (12 - n)/7 ] = -1
12- n = -7
-n = -19
n = 19
Check the picture below.
![\stackrel{\textit{\Large Areas}}{\stackrel{triangle}{\cfrac{1}{2}(6)(6)}~~ + ~~\stackrel{semi-circle}{\cfrac{1}{2}\pi (3)^2}}\implies \boxed{18+4.5\pi} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{pythagorean~theorem}{CA^2 = AB^2 + BC^2\implies} CA=\sqrt{AB^2 + BC^2} \\\\\\ CA=\sqrt{6^2+6^2}\implies CA=\sqrt{6^2(1+1)}\implies CA=6\sqrt{2} \\\\\\ \stackrel{\textit{\Large Perimeters}}{\stackrel{triangle}{(6+6\sqrt{2})}~~ + ~~\stackrel{semi-circle}{\cfrac{1}{2}2\pi (3)}}\implies \boxed{6+6\sqrt{2}+3\pi}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cstackrel%7Btriangle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%286%29%286%29%7D~~%20%2B%20~~%5Cstackrel%7Bsemi-circle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20%283%29%5E2%7D%7D%5Cimplies%20%5Cboxed%7B18%2B4.5%5Cpi%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7Bpythagorean~theorem%7D%7BCA%5E2%20%3D%20AB%5E2%20%2B%20BC%5E2%5Cimplies%7D%20CA%3D%5Csqrt%7BAB%5E2%20%2B%20BC%5E2%7D%20%5C%5C%5C%5C%5C%5C%20CA%3D%5Csqrt%7B6%5E2%2B6%5E2%7D%5Cimplies%20CA%3D%5Csqrt%7B6%5E2%281%2B1%29%7D%5Cimplies%20CA%3D6%5Csqrt%7B2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Perimeters%7D%7D%7B%5Cstackrel%7Btriangle%7D%7B%286%2B6%5Csqrt%7B2%7D%29%7D~~%20%2B%20~~%5Cstackrel%7Bsemi-circle%7D%7B%5Ccfrac%7B1%7D%7B2%7D2%5Cpi%20%283%29%7D%7D%5Cimplies%20%5Cboxed%7B6%2B6%5Csqrt%7B2%7D%2B3%5Cpi%7D)
notice that for the perimeter we didn't include the segment BC, because the perimeter of a figure is simply the outer borders.
Answer: 2
Step-by-step explanation:
Given
The polynomial is 
is divided by 
According to the remainder theroem
Put 
in the polynomial to get the remainder
So, we get the remainder 2.
