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Crazy boy [7]
2 years ago
5

Use absolute value to express the distance between −13 and 17 on the number.

Mathematics
1 answer:
motikmotik2 years ago
8 0

Answer:

30

Step-by-step explanation:

|-13| = 13

13+17=30

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Given f(x)=x+7 and g(x)=x-3 what is f(x)+g(x) ? A.2x+4 B.10x C.2x-7 D.-2x+10
Zinaida [17]
F(x)+g(x) = (x+7) + (x-3) =
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3 years ago
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Line CD contains points C(7, 4) and D(6, 1). Line JK is parallel to line CD. Which two points lie on line JK?
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3 years ago
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At a train station, freight trains depart at a rate of 2 trains every 30 minutes and passenger trains depart at a rate of 1 trai
never [62]
In my work I used t=trains and m=minutes:
30+30= 60m/2t
<span>30+30= 60m/2t
</span><span>30+30= 60m/2t
</span><span>30+30= 60m/2t
</span><span>30+30= 60m/2t
</span><span>30+30= 60m/2t
</span><span>30+30= 60m/2t
</span><span>30+30= 60m/2t
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</span><span>30+30= 60m/2t
</span><span>30+30= 60m/2t
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6 0
3 years ago
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:
kondaur [170]
\text{Proof by induction:}
\text{Test that the statement holds or n = 1}

LHS = (3 - 2)^{2} = 1
RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS
\text{Thus, the statement holds for the base case.}

\text{Assume the statement holds for some arbitrary term, n= k}
1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}

\text{Prove it is true for n = k + 1}
RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}

LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}
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= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}
= \frac{k(6k^{2} + 15k + 11) + 2}{}
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= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}
= RHS

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.
3 0
3 years ago
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