I believe that the answer is C because lets say you decide that the cube will roll on 1 six times. if you roll the cube 6 times, then there is a 1/6 probability of the cube landing on 1, but the more times you roll the cube, the more chances there will be of the number landing that many times as you theorized
thanks for free pints.........
Answer:
y=-4x-4
Step-by-step explanation:
y=-4x-4
y=mx+b where m=slope and b=y-intercept.
Answer:
a= 24b/5c b= 5ac/24 c= 24b/5a
Step-by-step explanation:
Put the problem into a calculator that can solve with variables.
Answer:
(2x-1)(2x+1)(x^2+2) = 0
Step-by-step explanation:
Here's a trick: Use a temporary substitution for x^2. Let p = x^2. Then 4x^4+7x^2-2=0 becomes 4p^2 + 7p - 2 = 0.
Find p using the quadratic formula: a = 4, b = 7 and c = -2. Then the discriminant is b^2-4ac, or (7)^2-4(4)(-2), or 49+32, or 81.
Then the roots are:
-7 plus or minus √81
p= --------------------------------
8
p = 2/8 = 1/4 and p = -16/8 = -2.
Recalling that p = x^2, we let p = x^2 = 1/4, finding that x = plus or minus 1/2. We cannot do quite the same thing with the factor p= -2 because the roots would be complex.
If x = 1/2 is a root, then 2x - 1 is a factor. If x = -1/2 is a root, then 2x+1 is a factor.
Let's multiply these two factors, (2x-1) and (2x+1), together, obtaining 4x^2 - 1. Let's divide this 4x^2 - 1 into 4x^4+7x^2-2=0. We get x^2+2 as quotient.
Then, 4x^4+7x^2-2=0 in factored form, is (2x-1)(2x+1)(x^2+2) = 0.
