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2 years ago

What is the m∠q ?????????????????????? help i taking a test and imma get beat if i fail

Mathematics

2 answers:

balandron [24]2 years ago

5 0

Answer:

Unlike the previous problem, this one requires application of the Law of Cosines. You want to find angle Q when you know the lengths of all 3 sides of the triangle.

Law of Cosines: a^2 = b^2 + c^2 - 2bc cos A

Applying that here:

40^2 = 32^2 + 64^2 - 2(32)(64)cos Q

Do the math. Solve for cos Q, and then find Q in degrees and Q in radians.

Step-by-step explanation:

zysi [14]2 years ago

4 0

Answer:

x = 30

Step-by-step explanation:

this is an isosceles triangle, which mean that PRQ and PQR are equal.

all sides of a triangle equal 180.

using this, you can make the following equation:

x + (2x + 15) + (2x + 15) = 180

combine like terms

5x + 30 = 180

subtract 30 from both sides

5x = 150

divide both sides by 5

x = 30

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2 years ago

Use mathematical induction to prove that for each integer n > 4,5" > 2^2n+1 + 100.

Flura [38]

Answer:

The inequality that you have is $5^{n}>2^{2n+1}+100,\,n>4$ . You can use mathematical induction as follows:

Step-by-step explanation:

For $n=5$ we have:

$5^{5}=3125$

$2^{(2(5)+1)}+100=2148$

Hence, we have that $5^{5}>2^{(2(5)+1)}+100.$

Now suppose that the inequality holds for $n=k$ and let's proof that the same holds for $n=k+1$ . In fact,

$5^{k+1}=5^{k}\cdot 5>(2^{2k+1}+100)\cdot 5.$

Where the last inequality holds by the induction hypothesis.Then,

$5^{k+1}>(2^{2k+1}+100)\cdot (4+1)$

$5^{k+1}>2^{2k+1}\cdot 4+100\cdot 4+2^{2k+1}+100$

$5^{k+1}>2^{2k+3}+100\cdot 4$

$5^{k+1}>2^{2(k+1)+1}+100$

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2 years ago

The inequality of the graph

fiasKO [112]

Answer:

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Step-by-step explanation:

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