These questions can be daunting at first, but they're pretty simple to solve.
First, we need to establish a common denominator. We have 2 / 3, 1 / 4, and
-4 / 3. The least common denominator we can get is by multiplying 4 and 3 together to get 12. So we will change the denominator as follows;
2 / 3, 1 / 4, -4 / 3 = 8 / 12, 3 / 12, -16 / 12
Now we can put these back into the equation.
8/12x + 3/12 = -16/12
8x + 3 = -16
It's simple math from here on out, but I'll show the process. What we can basically do now is take away the denominator because it doesn't matter now that it's common.
Subtract 3 from both sides. Now we have 8x = -19
Dividing by 8 on both sides of the equation will get you your answer.
x = -19/8
Hope this helps!
There are eight books total and he only grabbed two so the probability would be 2 out of 16 or 2:16
Answer: C 28.47° latitude and -7.35° longitude
Step-by-step explanation:
Answer:
0.75 mg
Step-by-step explanation:
From the question given above the following data were obtained:
Original amount (N₀) = 1.5 mg
Half-life (t₁/₂) = 6 years
Time (t) = 6 years
Amount remaining (N) =.?
Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:
Half-life (t₁/₂) = 6 years
Time (t) = 6 years
Number of half-lives (n) =?
n = t / t₁/₂
n = 6/6
n = 1
Finally, we shall determine the amount of the sample remaining after 6 years (i.e 1 half-life) as follow:
Original amount (N₀) = 1.5 mg
Half-life (t₁/₂) = 6 years
Number of half-lives (n) = 1
Amount remaining (N) =.?
N = 1/2ⁿ × N₀
N = 1/2¹ × 1.5
N = 1/2 × 1.5
N = 0.5 × 1.5
N = 0.75 mg
Thus, 0.75 mg of the sample is remaining.
So we can start with the full of possibilities and eliminate them one by one.
The full set is {0,1,2,3,4,5,6,7,8,9}.
Now we know that any prime greater than 2 is odd as otherwise it would have 2 as a factor, so we can eliminate all of these digits that would be an even number, leaving:
{1,3,5,7,9}
We also know that any prime greater than 5 cannot be a multiple of 5 and that all numbers with 5 in the digits are a multiple of 5, so we can eliminate 5.
{1,3,7,9}
We know that 11,13,17 and 19 are all primes, so we cannot eliminate any more of these, leaving the set:
{1,3,7,9} as our answer.
