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denis-greek [22]
3 years ago
8

Which side lengths form a right triangle?

Mathematics
1 answer:
3241004551 [841]3 years ago
3 0

Answer:  A

5,12,13 is a common Pythagorean triple, the second on the list after 3,4,5

B and C aren't on the list, and they would be if they were Pythagorean Triples, which is a list of natural number length sides (a,b,c)  which satisfy the Pythagorean Theorem, a²+b²=c², making them the sides of a right triangle.

Let's verify:

5²+12² = 25+144 = 169 = 13²              √

4²+4² = 16+16 = 32 ≠ 8²                     √

2²+3² = 4+9 = 13 ≠ 4²                         √

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F(x) = x^2 - 6x + 4               {-3, 0, 5}   (x, y) x is the domain, y is the range.

plug in the domain numbers into the equation to find the range.

y = x^2 - 6x + 4
y = -3^2 - 6(-3) + 4
y = 9 - (-18) + 4
y = 9 + 18 + 4
y = 31                        (-3, 31)

y = x^2 - 6x + 4
y = 0^2 - 6(0) + 4
y = 0 - 0 + 4
y = 4                       (0, 4)

y = x^2 - 6x + 4
y = 5^2 - 6(5) + 4
y = 25 - 30 + 4
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c. {-1, 4, 31}

hope this helped, God bless!
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Evaulate the function at the indicated values h(t)=t+(3/t)i find h(x-1)
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Note that the domain of h is <span>[0,∞]</span>.

By differentiating,

<span>h'<span>(t)</span>=<span>34</span><span>t<span>−<span>14</span></span></span>−<span>34</span><span>t<span>−<span>34</span></span></span></span>

by factoring out <span>34</span>,

<span>=<span>34</span><span>(<span>1<span>t<span>14</span></span></span>−<span>1<span>t<span>34</span></span></span>)</span></span>

by finding the common denominator,

<span>=<span>34</span><span><span><span>t<span>12</span></span>−1</span><span>t<span>34</span></span></span>=0</span>

<span>⇒<span>t<span>12</span></span>=1⇒t=1</span>

Since <span>h'<span>(0)</span></span> is undefined, <span>t=0</span> is also a critical number.

Hence, the critical numbers are <span>t=0,1</span>.

I hope that this was helpful.

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