The zeroes of <span>f(x) = x^5-12x^2+32x can be found by factoring,
</span><span>f(x) = x^5-12x^2+32x=(x-8)(x-4)
By the zero product theorem, (x-8)=0 or (x-4)=0 which means
x=8 or x=4.
So the zeroes of f(x) are S={4,8}</span>
Answer:
Line up the numbers on the right - do not align the decimal points.
Starting on the right, multiply each digit in the top number by each digit in the bottom number, just as with whole numbers.
Add the products.
Step-by-step explanation:
The answer is B) 0 miles. Because displacement is the shortest distance between starting point and ending point. Since, his starting point and ending point is the same, hence, his displacement is 0 miles.
Answer:
15 ft
Step-by-step explanation:
In a parallelogram the diagonals bisects each other
So WC= CY
solve the equation for x
Subtract x from both sides
Now we add 7 on both sides
x=11
Now we need to find CY
Plug in 11 for x
To factor out you have to think what multiples to AC and adds to B.
Ax^2+Bx+C
So... for this problem AxC=1x-24 or -24
B is -2.
So what two numbers multiply to -24: -3x8, -8x3, -4x6, -6x4, -2x12, -12x2.
Out of these, which adds to -2: -6+4=-2.
So the factors are (d-6)(d+4)
OR the longer way, which you really only use if A is not equal to 1.
Use the terms above and then rewrite the equation with two middle terms: d^2+4d-6d-24
Group the terms by using addition: (d^2+4d)+(6d-24)
Find what they have in common and factor it out. For the first, it's d. They both have d. So: d(d+4)
To check this, distribute the d. It should equal the first set lf parenthesis.
For the second, they have a number in common. 6 is a multiple of 24 so you can take that out: -6(d+4)
If the terms inside the parenthesis are the same, that's good. It means we can pair the insides and the outsides together to form the factors.
The two terms outside the parenthesis: d, -6 group together and become (d-6)
The inside terms stay the same: (d+4)
(d-6)(d+4)
Again, this is the longer way and no necessary for a problem like this. But if it was 2d^2, then this would be perfecf.
