Answer:
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Step-by-step explanation:
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The point G on AB such that the ratio of AG to GB is 3:2 is; G(4.2, 2)
How to partition a Line segment?
The formula to partition a line segment in the ratio a:b is;
(x, y) = [(bx1 + ax2)/(a + b)], [(by1 + ay2)/(a + b)]
We want to find point G on AB such that the ratio of AG to GB is 3:2.
From the graph, the coordinates of the points A and B are;
A(3, 5) and B(5, 0)
Thus, coordinates of point G that divides the line AB in the ratio of 3:2 is;
G(x, y) = [(2 * 3 + 3 * 5)/(2 + 3)], [(2 * 5 + 3 * 0)/(2 + 3)]
G(x, y) = (21/5, 10/5)
G(x, y) = (4.2, 2)
Read more about Line segment partition at; brainly.com/question/17374569
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Answer:
I would need more information to help you
Step-by-step explanation:
Answer:
3
Step-by-step explanation:
Since
3 : 2 = X : 2
Then we know
3/2 = X/2
Multiplying both sides by 2 cancels on the right
2 × (3/2) = (X/2) × 2
2 × (3/2) = X
Then solving for X
X = 2 × (3/2)
X = 3
Therefore
3 : 2 = 3 : 2
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:
n!/((n-r)!r!)
In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.
5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5
5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10
5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10
5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5
Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.
Now we add together the combinations
1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.
The answer is 32.
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Note: There is also an equation for permutations which is:
n!/(n-r)!
Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.
We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.
I hope this helps! If you have any questions, let me know :)
