7. 21.39
8. 6
those are the only two i know
We are given the functions:
<span>S (p) = 40 + 0.008 p^3 --->
1</span>
<span>D (p) = 200 – 0.16 p^2 --->
2</span>
T o find for the price in which the price of supply equals
demand, all we have to do is to equate the two equations, equation 1 and 2, and
calculate for the value of p, therefore:
S (p) = D (p)
40 + 0.008 p^3 = 200 – 0.16 p^2
0.008 p^3 + 0.16 p^2 = 160
p^3 + 20 p^2 = 20,000
p^3 + 20 p^2 – 20,000 = 0
Calculating for the roots using the calculator gives us:
p = 21.86, -20.93±21.84i
Since price cannot be imaginary therefore:
p = 21.86
Answer:
Step-by-step explanation:
In a G.P, the nth term is given as
Un=ar^(n-1)
Where
a is first term
n is nth term
And r is common ratio
So in the question given above,
The third term exceed the first term by 16
i.e U3=U1+16
Where U1=a. First term
U3=a+16
Given also that, the sum if the third term and fourth term is 72.
Then U3+U4=72.
We are told to find common ratio (r)
U3=ar^3-1
U3=ar^2
Also, U4=ar^3
U3+U4=72
ar^2+ar^3=72
ar^2(1+r)=72. equation 1
Also for
U3=a+16
ar^2=a+16
ar^2-a=16
a(r^2-1)=16. From (x^2-y^2)=(x+y)(x-y)
Then,
a(r-1)(r+1)=16. Equation 2
Divide equation 2 by equation 1
a(r-1)(r+1)/ar^2(1+r) =16/72
Then a cancel a and (1+r) cancel (1+r)
So,
(r-1)/r^2=2/9
Cross multiply
9(r-1)=2r^2
2r^2-9r+9=0
Solving the quadratic equation
2r^2-6r-3r+9=0
2r(r-3)-3(r-3)=0
(r-3)(2r-3)=0
r-3=0. Or. 2r-3= 0
Then r=3 or r=3/2
Answer:
y=5/2x+5
plz mark me as brainliest
Answer: f(x) · g(x)=-32x³+104x²-100x+100
Step-by-step explanation:
Since we are given f(x) and g(x), all we have to do is multiply them together.
f(x) · g(x)=(4x-10)(-8x²+6x-10) [distribute by FOIL]
f(x) · g(x)=-32x³+24x²-40x+80x²-60x+100 [combine like terms]
f(x) · g(x)=-32x³+104x²-100x+100
Now, we have multiplied them together, we have f(x) · g(x)=-32x³+104x²-100x+100.
