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Yuri [45]
3 years ago
11

Directions for constructing. Orthocenter of a triangle​

Mathematics
1 answer:
ioda3 years ago
3 0

Answer:

Step-by-step explanation:

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Answers are C and F.....
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What is the definition of csc θ ?
Greeley [361]
Csc is the inverse of sin. This means the formula in regards to a triangle for csc is h/o, unlike o/h for sin. csc can also be knows as 1/sin. 
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Look at the stem-and-leaf plot.
cestrela7 [59]
The answer is c 10 #s are greater than 110
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2 years ago
What does 3ˣ + 4·3ˣ⁺¹ =<br> a. 13·3ˣ⁺¹<br> b. 5·3ˣ⁺¹<br> c. 5·3²ˣ⁺¹<br> d. 13·3ˣ
Crazy boy [7]

Answer:

D. 13*3^x

Step-by-step explanation:

3^x +4*3^x^+^1= \\3^x+4*3(3^x)=\\3^x(1+[4*3])=\\3^x(1+12)=\\3^x(13)=\\13*3^x

3 0
3 years ago
Mathematical induction, prove the following two statements are true
adelina 88 [10]
Prove:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}
____________________________________________

Base Step: For n=1:
n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1
and
4-\dfrac{n+2}{2^{n-1}}=4-3=1
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}
assumed to be true.

--------------------------------------------------------------------------

Induction Step: For n=k+1:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}

Distribute the -2 and combine the fractions together,
=4+\dfrac{-2k-4+(k+1)}{2^{k}}

Combine like-terms,
=4+\dfrac{-k-3}{2^{k}}

pull the negative back out,
=4-\dfrac{k+3}{2^{k}}

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}
3 0
2 years ago
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