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Ymorist [56]
3 years ago
10

How to find the sum of 2+4+6....+40?

Mathematics
1 answer:
grandymaker [24]3 years ago
3 0
Answer 420.
a, is the first term. d is difference.
Since this is an arithmetic progression, I use this formula.
Un = nth term
Sn = sum of terms.
First find the number of terms, which is 20.
Then use the second formula to find the sum.

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Simplify the following expression by combining like terms.
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Answer: the answer is A.

Step-by-step explanation:

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The perimeter of a triangle is 41 inches. The three sides of the triangle are all differnt lenghts. The longest side is 10 less
erma4kov [3.2K]

Answer:

Longest\ side=20\ inches\\\\Middle\ side=15\ inches\\\\Shortest\ side=6\ inches

Step-by-step explanation:

Let be "x" the lenght of the longest side, "y" the lenght of the middle side and "z" the lenght of the shortest side.

The perimeter is:

41=x+y+z       [Equation 1]

We know that the longest side is 10 less than twice the middle side length. This can be expressed as:

x=2y-10          [Equation 2]

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y=3z-3         [Equation 3]

The steps are:

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y=3z-3\\\\y+3=3z

z=\frac{y+3}{3}    [Equation 4]

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- Substitute [Equation 2] into [Equation 1]

Then:

41=(2y-10)+y+(\frac{y+3}{3})

- Solve for "y":

41=2y-10+y+\frac{y+3}{3}\\\\41=2y-10+y+\frac{y}{3}+1\\\\41=\frac{10}{3}y-9\\\\41+9=\frac{10}{3}y\\\\\frac{50*3}{10}=y\\\\y=\frac{150}{10}\\\\y=15

- Substitute this value into [Equation 2] and into [Equation 4]:

x=2(15)-10\\\\x=20

z=\frac{(15)+3}{3}\\\\z=6

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</span>Δ Δ Δ Δ Δ
Δ Δ Δ Δ Δ

So I want to take away 3 triangles, then by taking away 3<em> what is left? </em>Well the answer is 7:

Δ Δ Δ Δ Δ
Δ Δ 

That is, if we subtract 3 from 10 then the result is:

10-3=7

Second way. In this subtraction we take the same 10 triangles:

Δ Δ Δ Δ Δ
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