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4 years ago

How to solve this triangle ABC?i did super wrong all :( who can help me please?

Mathematics

1 answer:

Alex73 [517]4 years ago

5 0

It looks like you had some of the numbers correct, but you just labeled it and rounded it incorrectly, and the teacher gave you super partial credit for it.

For my work, I'm assuming that side A is opposite of angle A and side B is opposite of angle B, etc...

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karen is reading the great gatsby in English class. So far, she has read 30 pages. If she reads 18 pages a day and there are 192

Brums [2.3K]

Answer:

it will take her 9 days

Step-by-step explanation:

192- 30 = 162

162 / 18 = 9

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3 years ago

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Can someone please help me

s344n2d4d5 [400]

the answer is 2p - 4 :)

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3 years ago

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Yoo can some on help im just a dud that suck a ixl

Airida [17]

I think the answer is 58, but im not 100% sure so im sorry if it is wrong

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3 years ago

You roll a fair 6-sided die. What is P(roll less than 4)?

frutty [35]

Answer:

1/2

Step-by-step explanation:

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3 years ago

Given that 'n' is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction

Oliga [24]

The base case is the claim that

$\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}$

which reduces to

$\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86$

which is true.

Assume that the inequality holds for n = k ; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}$

We want to show if this is true, then the equality also holds for n = k + 1 ; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}$

By the induction hypothesis,

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}$

Now compare this to the upper bound we seek:

$\dfrac{2k+1}{k+1} > \dfrac{2k+2}{k+2}$

because

$(2k+1)(k+2) > (2k+2)(k+1)$

in turn because

$2k^2 + 5k + 2 > 2k^2 + 4k + 2 \iff k > 0$

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3 years ago

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