Question 58 a through f.
1 answer:
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Answer:
<h2>There are 13 quarters, 5 nickels and 20 dimes.</h2>Step-by-step explanation:
Givens
- Max has $5.50.
- He has eight more quarters than he does nickels.
- He had four times as many dimes as he has nickels.
From these statements, we can deduce algebraical expressions to solve the problem. Let's call quarters,
nickels and
dimes.
(8 more quarters than nickels)
(he has four times as many dimes as he has nickels).
(Max has a total of $5.50, one quartes is $0.25, one nickel is $0.05 and one dime is $0.10).
Let's replace the first two expressions into the third equation
There are 5 nickels.
Now, we use this value to find the number of quarters
There are 13 quarters.
Also,
There are 20 dimes.
Therefore, there are 13 quarters, 5 nickels and 20 dimes.
<h3>Answer: 40</h3>
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Explanation:
JQ is longer than QN. We can see this visually, but the rule for something like this is the segment from the vertex to the centroid is longer compared to the segment that spans from the centroid to the midpoint.
See the diagram below.
The ratio of these two lengths is 2:1, meaning that JQ is twice as long compared to QN. This is one property of the segments that form when we construct the centroid (recall that the centroid is the intersection of the medians)
We know that JN = 60
Let x = JQ and y = QN
The ratio of x to y is x/y and this is 2/1
x/y = 2/1
1*x = y*2
x = 2y
Now use the segment addition postulate
JQ + QN = JN
x + y = 60
2y + y = 60
3y = 60
y = 60/3
y = 20
QN = 20
JQ = 2*y = 2*QN = 2*20 = 40
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We have
JQ = 40 and QN = 20
We see that JQ is twice as larger as QN and that JQ + QN is equal to 60.