<span> Given polynomial x^2+8x-48 = 0</span>
<span>x^2+12x-4x-48 = 0</span>
<span>x(x+12)-4(x+12) = 0</span>
<span>(x+12)(x-4) = 0</span>
<span>x+12 = 0</span>
Subtract 12 from each side.
<span>x+12-12 = 0-12</span>
<span>x = -12</span>
<span>and x-4 = 0</span>
Add 4 to each side.
<span>x-4+4 = 0+4</span>
<span>x = 4</span>
<span>Roots are -12,4.</span>
Addition: 70 + 80 = 150. You round 65 up because the ones digit is greater than and the same with 77.
Multiplication: Round up since both ones digits are bigger than 5. 70 x 80= 5600.
Answer:
None of these.
Step-by-step explanation:
Let's assume we are trying to figure out if (x-6) is a factor. We got the quotient (x^2+6) and the remainder 13 according to the problem. So we know (x-6) is not a factor because the remainder wasn't zero.
Let's assume we are trying to figure out if (x^2+6) is a factor. The quotient is (x-6) and the remainder is 13 according to the problem. So we know (x^2+6) is not a factor because the remainder wasn't zero.
In order for 13 to be a factor of P, all the terms of P must be divisible by 13. That just means you can reduce it to a form that is not a fraction.
If we look at the first term x^3 and we divide it by 13 we get
we cannot reduce it so it is not a fraction so 13 is not a factor of P
None of these is the right option.
Answer:
The answer is 28/9
Step-by-step explanation:
Step one
Given information
We are given that the function is

And we are expected to find f(6), that is substitute the value of x with 6
Step two:

Function A is linear
y = 5/3x + 0
let me know if this is incorrect