X^2=x+12
minus x+12 both sides
x^2-x-12=0
factor
(x-3)(x+4)=0
set each to zero
x-3=0
x=3
x+4=0
x=-4
x=3 or -4
answer is first one
F(t) = 2(0.4)^t
To solve f(-2) we should substitute t = -2 into the equation:
f(-2) = 2(0.4)^(-2)
= 2(6.25)
= 12.5
1) 6^(3x)=104
loga =logarithm to the base a
loga n=z ⇔a^z=n
Therefore:
log6 104=3x
x=(log6 104) / 3
x=(ln 104) / 3ln 6=0.8640279....≈0.864 (loga z=ln a / ln z)
Answer: x≈0.864
2) log3 (x+2)+ log3 (4x-1)=5
log3 x=5 ⇔ 3⁵=x ⇒x=243
log3 [(x+2)(4x-1)]=log3 243
log3 (4x²+7x-2)=log3 243
Therefore:
4x²+7x-2=243
4x²+7x-245=0
we solve this square equation:
x=[-7⁺₋√(49+3920)] / 8 =(-7⁺₋63) / 8
we have two solutions:
x₁=(-7-63) / 8=-8.75 this solution is not valid.
x₂=(-7+63) /8=7
Answer: x=7
Answer: C
Step-by-step explanation:
Answer:
Step-by-step explanation:
There are non-distinct sums that can be achieved when rolling two fair sided dice.
The smallest of these sums is and the largest of these sums is . Within this range, there exists only one perfect cube, .
Count how many ways we can achieve a sum of 8 with two dice:
Thus the probability the total score (sum) will be a perfect cube when rolling two fair six-sided dice is equal to