<span>first find the similar term, the similar term is x^2y.
10x^2y + 2xy^2- 4x^2y
group together the similar term and place them in a parenthesis
then perform the operation to simplify the expression.
= </span>(10x^2y<span><span> - 4x^2y
) </span>+ 2xy^2
= ( </span><span>6x^2y ) + </span><span>2xy^2
= </span><span>6x^2y </span><span><span>+ <span>2xy^2 is the sum of the polynomial</span></span> </span>
Answer:
12.5
Step-by-step explanation:
We can use ratios to solve
4 4+6
---- = ----------
5 AC
4 10
---- = ----------
5 AC
Using cross products
4 * AC = 5*10
4AC = 50
Divide by 4
AC = 50/4
AC =12.5
$16.8. the tip would end up being $2.8 and you add that to the original $14
Answer:
Step-by-step explanation:
Put these numbers into their prime factors.
18: 2 * 3 * 3
27: 3 * 3 * 3
12: 2 * 2 * 3
The LCM must have
two 2s
Three 3s
That's it
2*2 * 3 * 3 * 3
LCM = 108
648 might be a multiple of the three, but it's not the smallest one.
108 = 27 * 4
108 = 18 * 6
108 = 12 * 9
<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.