Answer:
D. Rx) = x2 - 4x + 10
Step-by-step explanation:
R(x)=(x-2)^2+6
R(x)=(x-2)^2+6
=(x-2)(x-2)+6
=x^2-2x-2x+4+6
=x^2-4x+10
R(x) = x^2 - 4x + 10
Option D is the answer
Its <span>If Kevin and Amanda continue to train until week 16, what will their times be? 6. Do you believe a linear model best represents the relationship of the time of the runners and the weeks that passed?(Hint: look at question 5). What do you think this says about problems in the real world? Justify your thoughts in 3-4 sentences. </span> cause they are talking about minutes and per miles
Answer:
-7/3
Step-by-step explanation:
2(6−4)=3(6+2)
2(6x-4)=3(6x+2)
Solve
1
Distribute
2(6−4)=3(6+2)
{\color{#c92786}{2(6x-4)}}=3(6x+2)
12−8=3(6+2)
{\color{#c92786}{12x-8}}=3(6x+2)
2
Distribute
12−8=3(6+2)
12x-8={\color{#c92786}{3(6x+2)}}
12−8=18+6
12x-8={\color{#c92786}{18x+6}}
3
Add
8
8
to both sides of the equation
12−8=18+6
12x-8=18x+6
12−8+8=18+6+8
12x-8+{\color{#c92786}{8}}=18x+6+{\color{#c92786}{8}}
5 more steps
Solution
=−7/3
Divide across and just bring down the numbers
Putting this as an arithmetic sequence gives:
The sum of the series = 16 x 7 x 7 = 784 m^3 = 784 000 L
The sum of an arithmetic series can be written as:
![S_n=n/2 [2a+(n-1)d] = 784 000 \\n/2[2(150)+(n-1)200] = 784 000 \\n[300+200(n-1)=1 568 000 \\300n+200n^2-200n = 1 568 000 \\200n^2+100n- 1 568 000 = 0 \\2n^2 +n- 15680 = 0 \\n= 88.2...,-88.7](https://tex.z-dn.net/?f=S_n%3Dn%2F2%20%5B2a%2B%28n-1%29d%5D%20%3D%20784%20000%0A%5C%5Cn%2F2%5B2%28150%29%2B%28n-1%29200%5D%20%3D%20784%20000%0A%5C%5Cn%5B300%2B200%28n-1%29%3D1%20568%20000%0A%5C%5C300n%2B200n%5E2-200n%20%3D%201%20568%20000%0A%5C%5C200n%5E2%2B100n-%201%20568%20000%20%3D%200%0A%5C%5C2n%5E2%20%2Bn-%2015680%20%3D%200%0A%0A%5C%5Cn%3D%2088.2...%2C-88.7)
n has to be positive, so we get
n =
<u>88.2 hours (3 s.f.)</u>
