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Cerrena [4.2K]
2 years ago
12

I need some serious help with this asap!!! I will mark brainliest if right! Explain

Mathematics
1 answer:
34kurt2 years ago
6 0

Answer:

I'm sorry this isn't the answer, but I just want you to know that you are incredible and that I love you for you! You are special to everyone you meet, and should not change who you are. I know your life may be tough, but you are strong and can get through it!

Step-by-step explanation:

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Solve 5n – 7p + 3n = 25p for n.
saul85 [17]

5n – 7p + 3n = 25p

Combine like terms

8n - 7p = 25p

Add (7p) to both sides

8n - 7p + 7p = 25p + 7p

Simplifying

8n = 32p

Divide both sides by 8

8n / 8 = 32p / 8

Simplifying

n = 4p

7 0
3 years ago
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I need help asap :):):)
KiRa [710]

Answer:

24 ÷ 4

Step-by-step explanation:

Putting something into fourths is putting into 4 parts. If you divide 24 by 4 you get 6. 6 equals 1/4 of 24. To check your work add 6 together 4 times.

7 0
3 years ago
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Approximately what portion of the circle is shaded blue?<br><br> A. 2/3<br> B. 2/10 C.2/5
Nuetrik [128]
C) closest to 2/5.
2/3 is bigger than half 2/10 would be like less than 1/4
3 0
2 years ago
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What is 1+1 please please please
mojhsa [17]
The answer to this question is 2
6 0
2 years ago
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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 <em>n</em> = <em>k </em>; that

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

We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 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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2 years ago
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