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2 years ago

For any real number a,

="\sqrt{a^{2} }" align="absmiddle" class="latex-formula">=

Mathematics

1 answer:

Mrac [35]2 years ago

6 0

Step-by-step explanation:

for any real number a, the definition is:

$\sqrt{ {a}^{2} } = |a|$

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What is -8 +4 using a number line

vladimir1956 [14]

Answer:

-4

Step-by-step explanation

start on -8 in the number line and move to the right 4 times, leaving you at -4

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3 years ago

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What is four to the second power+9-2 In numeric expression

Rufina [12.5K]

Answer:

Step-by-step explanation:

4^2+9-2

so use PEMDAS

so do the 4^2 first which is the same as 4•4=16

16+9-2

then the adding so 16+9

25-2 then subtract

23 is your answer

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4 years ago

A quantity, y, varies directly as x. When y = 9, x = 4. Find y when x = 10

Kipish [7]

Y = kx

9 = k4

k = (9/4)

y = (9/4)x

when x=10,

y = (9/4)10
y = 22.5

5 0

4 years ago

Given that 7 x − 2 y = 35 Find y when x = − 9

Sergio [31]

so x = minus 9

then

7 × (-9) - 2y = 35

-63 - 2y = 35

-2y = 35 plus 63

-2y = 98

- y = 49

y = -49

3 0

3 years ago

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Prouvez par récurrence que quel que soit n EN\{0}, on a

Vanyuwa [196]

The left side is equivalent to

$\displaystyle \sum_{k=1}^n \frac1{k(k+1)}$

When n = 1, we have on the left side

$\displaystyle \sum_{k=1}^1 \frac1{k(k+1)} = \frac1{1\cdot2} = \frac12$

and on the right side,

$1 - \dfrac1{1+1} = 1 - \dfrac12 = \dfrac12$

so this case holds.

Assume the equality holds for n = N, so that

$\displaystyle \sum_{k=1}^N \frac1{k(k+1)} =1 - \frac1{N+1}$

We want to use this to establish equality for n = N + 1, so that

$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac1{N+2}$

We have

$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = \sum_{k=1}^N \frac1{k(k+1)} + \frac1{(N+1)(N+2)}$

$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac1{N+1} + \frac1{(N+1)(N+2)}$

$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac{N+2}{(N+1)(N+2)} + \frac1{(N+1)(N+2)}$

$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac{N+1}{(N+1)(N+2)}$

$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac1{N+2}$

and this proves the claim.

6 0

3 years ago