Answer: x=-17
Step-by-step explanation: Divide by -1 then add 8 to both sides and get (x+9)^3=-512, and the cube root of -512 is -8, so x+9=-8, and subtracting 9 from both sides we get x=-17 as answer.
Answer:How many fives are in 15? ... How many three-fourths are in 2? ... There are 6 half-sized pieces in 3 wholes. ... and we can see that there are 4 wholes with 6 sixths in each whole, so there are 4\times 6 = 24 sixths in 4. ... Each whole yields a two-thirds and one half of another two-thirds, therefore 3 sets of two-thirds can be ...
Step-by-step explanation:
Answer:
The factors are (5x + 3) and (2x + 1)
Step-by-step explanation:
When you need to factor a quadratic, and the coefficient of the x² is not 1, use the slide and divide method.
The general form of a quadratic is ax² + bx + c
Factor: 10x² + 11x + 3
Here a = 10, b = 11, and c = 3
Step 1: Multiply ac, we SLIDE a over to c. Notice the 10 is gone for now..
x² + 11x + 30
Step 2: Factor this (this step will always factor)
x² + 11x + 30 = (x + 5)(x + 6)
So the factors are (x + 5)(x + 6), but we now need to DIVIDE by a, since we multiplied it into c before. We divide the constants in the factors...
(x + 5/10 )(x + 6/10 )
Now reduce the fractions as much as possible...
(x + 1/2 )(x + 3/5)
*If they don't reduce to a whole number, SLIDE the denominator over as a coefficient of x....
(2x + 1)(5x + 3) *2 slide over in front of x, 5 slide over in front of x, the fractions are gone!
These are our factors!
Answer:
A. Repeat the simulation several more times
Step-by-step explanation:
The purpose of the simulation model is to represent the effectivity of the passes.
The proportion of successful passes is 60%.
As we have 10 digits available, 6 (digits from 0 to 5) are used for the outcome "the pass is completed" and 4 (digits 6, 7, 8, and 9) to represent the outcome "the pass is not completed". This is correct, as it represents a probability of 60% of having a successful pass.
But to have a representative distribution of the possible and probable results, the simulation have to run enough times to have a stable distribution of the results.
