The roots are 1 +√7 and 1 -√7.
<h3>What is Quadratic equation?</h3>
A quadratic equation in the variable x is an equation of the form ax² + bx + c= 0, where a, b, c are real numbers, a≠0
Given equation:
y= x²+2x-6
First,
Half the coefficient of x and add and subtract the square of (b/2)
y= x²+2x-6+(1)²-(1)²
y= x²+2x+(1)² -6 -(1)²
y= (x+1)² -7
Now, equate y=0
(x+1)² -7 =0
(x+1)² = 7
x+1= ±√7
x=1 ±√7
Hence, the roots are 1 +√7 and 1 -√7.
Learn more about quadratic equation here:
brainly.com/question/1962219
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The correct answer is the 2nd choice because -1/6 is closer to 0 and the surface of the pool would be considered 0. -4/6 is farther away from 0 so it is less.
Answer:
3) f(x) = 3(x + 2)^2 - 1.
Step-by-step explanation:
f(x) = 3x² + 12x +11
f(x) = 3(x^2 + 4x) + 11 Completing the square:
f(x) = 3[(x + 2)^2 - 4] + 11
f(x) = 3(x + 2)^2 -+3*-4 + 11
f(x) = 3(x + 2)^2 - 12 + 11
f(x) = 3(x + 2)^2 - 1.
There's more than one way to combine them really
but an obvious one will be
![\bf \begin{array}{llll} h(x)&=&(f\circ g)(x)\\\\ &&\sqrt[3]{7x+1}\\\\ &&\sqrt[3]{g(x)}\leftarrow \begin{array}{llll} f(x)=\sqrt[3]{x }\\\\ g(x)=7x+1 \end{array} \end{array}\\\\ -----------------------------\\\\ (f\circ g)(x)\iff f[\quad g(x)\quad ]=\sqrt[3]{g(x)}\implies f[\quad g(x)\quad ]=\sqrt[3]{7x+1}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bllll%7D%0Ah%28x%29%26%3D%26%28f%5Ccirc%20g%29%28x%29%5C%5C%5C%5C%0A%26%26%5Csqrt%5B3%5D%7B7x%2B1%7D%5C%5C%5C%5C%0A%26%26%5Csqrt%5B3%5D%7Bg%28x%29%7D%5Cleftarrow%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Af%28x%29%3D%5Csqrt%5B3%5D%7Bx%20%7D%5C%5C%5C%5C%0Ag%28x%29%3D7x%2B1%0A%5Cend%7Barray%7D%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%28f%5Ccirc%20g%29%28x%29%5Ciff%20f%5B%5Cquad%20g%28x%29%5Cquad%20%5D%3D%5Csqrt%5B3%5D%7Bg%28x%29%7D%5Cimplies%20%20f%5B%5Cquad%20g%28x%29%5Cquad%20%5D%3D%5Csqrt%5B3%5D%7B7x%2B1%7D)
Answer:
$31.25
Step-by-step explanation:
125 divided by 4 = 31.25
$31.25
