Answer:
The ratio of George age to Carl's age is 1:12.
Step-by-step explanation:
Let the age of George be 'g'.
Let the age of Alex be 'a'.
Also Let the age of Carl be 'c'.
Given:
The sum of their ages is 68.
So equation can be framed as;
Also Given:
Alex is 12 years older than George.
So equation can be framed as;
Now Given:
Carl is three times older than Alex.
But 
So we get;
Now Substituting equation 2 and equation 3 in equation 1 we get;
Subtracting both side by 48 using subtraction property of equality we get;
Now Dividing both side by 5 using Division property of equality we get;
Hence George age 
Now Alex age 
Also Carl's age 
Now we need to find the ratio of George age to Carl's age.
Hence the ratio of George age to Carl's age is 1:12.
-- He must have at least one of each color in the case, so the first 3 of the 5 marbles in the case are blue-green-black.
Now the rest of the collection consists of
4 blue
4 green
2 black
and there's space for 2 more marbles in the case.
So the question really asks: "In how many ways can 2 marbles
be selected from 4 blue ones, 4 green ones, and 2 black ones ?"
-- Well, there are 10 marbles all together.
So the first one chosen can be any one of the 10,
and for each of those,
the second one can be any one of the remaining 9 .
Total number of ways to pick 2 out of the 10 = (10 x 9) = 90 ways.
-- BUT ... there are not nearly that many different combinations
to wind up with in the case.
The first of the two picks can be any one of the 3 colors,
and for each of those,
the second pick can also be any one of the 3 colors.
So there are actually only 9 distinguishable ways (ways that
you can tell apart) to pick the last two marbles.
We are given with the rate of calories burnt out through running of <span>100 calories per mile. If 10 miles is ran, we get the total calories burnt is </span><span>100 calories per mile * 10 miles equal to 1000 calories. Power (watts) is equal to Joules per second. 1000 calories is equal to 4185.8 Joiules. In an hour run- the power is 69.76 Watts</span>
The constant in this case is the number without the variable, making the answer A)3.
You know that three points A,B,CA,B,C (two vectors A⃗ BA→B, A⃗ CA→C) form a plane. If you want to show the fourth one DD is on the same plane, you have to show that it forms, with any of the other point already belonging to the plane, a vector belonging to the plane (for instance A⃗ DA→D).
Since the cross product of two vectors is normal to the plane formed by the two vectors (A⃗ B×A⃗ CA→B×A→C is normal to the plane ABCABC), you just have to prove your last vector A⃗ DA→D is normal to this cross product, hence the triple product that should be equal to 00:
A⃗ D⋅(A⃗ B×AC)=0
