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4 years ago

Solve equation with square roots k^2+6=6

Mathematics

1 answer:

il63 [147K]4 years ago

3 0

Answer:

k = 0

Step-by-step explanation:

Note the equal sign. What you do to one side, you do to the other. Do the opposite of PEMDAS. First, subtract 6 from both sides

k² + 6 (-6) = 6 (-6)

k² = 0

Isolate the k. Root both sides

√(k²) = √(0)

k = 0

0 is your answer for k

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Identify each expression that represents the slope of a tangent to the curve y=1/x+1 at any point (x,y) .

tankabanditka [31]

Answer:

Slope of a tangent to the curve = $f'(x) = \frac{-1 }{(x+1)^{2} }$

Step-by-step explanation:

Given - y = 1/x+1

To find - Identify each expression that represents the slope of a tangent to the curve y=1/x+1 at any point (x,y) .

Proof -

We know that,

Slope of tangent line = f'(x) = $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

We have,

f(x) = y = $\frac{1}{x+1}$

So,

f(x+h) = $\frac{1}{x+h+1}$

Now,

Slope = f'(x)

And

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\= \lim_{h \to 0} \frac{\frac{1}{x+h+1} - \frac{1}{x+1} }{h}\\= \lim_{h \to 0} \frac{x+1 - (x+h+1) }{h(x+1)(x+h+1)}\\= \lim_{h \to 0} \frac{x +1 - x-h-1 }{h(x+1)(x+h+1)}\\= \lim_{h \to 0} \frac{-h }{h(x+1)(x+h+1)}\\= \lim_{h \to 0} \frac{-1 }{(x+1)(x+h+1)}\\= \frac{-1 }{(x+1)(x+0+1)}\\= \frac{-1 }{(x+1)(x+1)}\\= \frac{-1 }{(x+1)^{2} }$

∴ we get

Slope of a tangent to the curve = $f'(x) = \frac{-1 }{(x+1)^{2} }$

6 0

3 years ago

What is 0.71 as a quotient of integers?

Blababa [14]

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So, you put 71 over 100.

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3 years ago

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Gnoma [55]

Answer:

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The number of ways to select 2 cards from 52 cards in case the order is important is 2652.

Step-by-step explanation:

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${n\choose k}=\frac{n!}{k!(n-k)!}$

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$^{n}P_{k}=\frac{n!}{(n-k)!}$