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soldi70 [24.7K]
3 years ago
10

Nicolas has three fewer than twice the number of songs downloaded as Sabrina does.

Mathematics
1 answer:
EleoNora [17]3 years ago
3 0
3 ajjajajajajajajajjaajja
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Pls help <br> Are they SSS or SAS??
8090 [49]

Answer:

I think they are SAS but I'm not sure

7 0
3 years ago
Don't understand please help with answer
Taya2010 [7]
((-4)^5)^8 / ((-4)^7)^3

= (-4)^40 / (-4)^21

= (-4)^20
4 0
3 years ago
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Find the surface area plz
vladimir1956 [14]
Hi again! ;)

If we want to find the surface area of a cube, we need to find the area of one surface first, and then multiply that by 9, since there are 9 sides/surfaces on a cube.

Since each side of a square is the same, we first find the area of one surface:

9 × 9 = 81

Then, to find the surface area, we multiply 81 by 6:

81 × 6 = 486

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486 \:  {in}^{2}
7 0
3 years ago
HELPPPPPP !!! I seriously don't know what to write.
tresset_1 [31]

Answer:

<em>This question was previously answered on Brainly...</em>

<em>Be sure to put this into your own words.</em>

<em>Answer: check explanation for the solution </em>

<em> </em>

Step-by-step explanation:

<em> </em>

A business that offers services to people by providing many amusement and fun with gate fees

The algebraic equations to be used is the general linear equation

Y = MX + C

Where

Y = total income or money realised

M = rate or price rate

X = number of goods or services

C = flat rate or gate fees

The business can also operate differently by using exponential equation

A = P(1 + R%)^t

Where

A = profit

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t = time

3 0
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 <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

6 0
2 years ago
Read 2 more answers
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