384|2
192|2
96|2
48|2
24|2
12|2
6|2
3|3
1
![384=2^7\cdot3=2^6\cdot2\cdot3=(2^2)^3\cdot6=4^3\cdot6\\\\ \sqrt[3]{384}=\\ \sqrt[3]{4^3\cdot6}=\\ \sqrt[3]{4^3}\cdot\sqrt[3]{6}=\\ 4\sqrt[3]6\Rightarrow C](https://tex.z-dn.net/?f=384%3D2%5E7%5Ccdot3%3D2%5E6%5Ccdot2%5Ccdot3%3D%282%5E2%29%5E3%5Ccdot6%3D4%5E3%5Ccdot6%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B384%7D%3D%5C%5C%0A%5Csqrt%5B3%5D%7B4%5E3%5Ccdot6%7D%3D%5C%5C%0A%5Csqrt%5B3%5D%7B4%5E3%7D%5Ccdot%5Csqrt%5B3%5D%7B6%7D%3D%5C%5C%0A4%5Csqrt%5B3%5D6%5CRightarrow%20C)
192|2
96|2
48|2
24|2
12|2
6|2
3|3
1
Evaluate the following-2167+179+(-128)-68
1 answer:
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Answer:
The equation of the parabola is y = 6-2b+1
Step-by-step explanation:
First we write a quadratic equation of parabola as y = a+bx+c
Now it is given in the question that parabola passes through three points (-5,161),(-2,29),(6,205).
To find the equation we have to get the value of a,b & c.
Now we form three equations to find the value of a,b,c.
We put the value (-5,161) in the equation
161 = a×25+b×(-5)+c
161 = 25a-5b+c----------(1)
Now we put (-2,29) in the equation
29 = a×4+b×(-2)+c
29 = 4a-2b+c---------(2)
Again we put (6,205) in the equation
205 = a×36+6b+c------------(3)
Now we subtract equation (1) from (2)
161-29 = 25a-4a-5b-(-2b)+c-c
132 = 21a-5b+2b
132 = 21a-3b
132 = 3(7a-b)
44 = 7a-b---------(4)
Now we subtract equation (2) from (3)
205-29 = 36a-4a+6b-(-2b)+c-c
176 = 32a+6b+2b
176 = 32a+8b
176 = 8(4a+b)
22 = 4a+b------------(5)
Now we add equation (5) and (4)
44+22 = 7a+4a-b+b
66 = 11a
a = 6
Now we put the value a in equation (4)
44 = 7×6-b
44 = 42-b
44-42 = -b
2 = (-b)
b = (-2)
Now we put the value of a and b in equation number (2)
29 = 4×6-2(-2)+c
29 = 24+4+c
29 = 28+c
c = 29-28
c =1
Finally we put the value of a,b,c in the quadratic equation
y = 6+(-2)x+1
y = 6-2+1
This is the final answer.