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2 years ago

Will give brainliest please helpppp

Mathematics

1 answer:

Mama L [17]2 years ago

7 0

Its not the first two, I’m having trouble seeing those bottom two and what they are.

3rd one: y=ln x

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Luden [163]

Your answer is A. True.

Hope this helps.

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2 years ago

Help me please!!!!!!!!

DedPeter [7]

I think it’s d for the standard equation

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3 years ago

Read 2 more answers

a lab uses negative numbers to describe a car moving backward on a road if the car was at -75.5 feet ,went forward 23 feet and t

never [62]

Did you that that God loves you and wants the best for you

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2 years ago

HELP PLEASE <br> 5/12-(x-3)/6≤(x-2)/3<br> Solve with a step by step solution

zlopas [31]

Answer:x>3 1/6

Step-by-step explanation:

5/12-(x-3)/6 <(x-2)/3

5/12 *12-(x-3)/6*12<(x-2)/6*12

5-2(x-3)<4(x-2)

5-2x+6<4x-8

-2x+11<4x-8

-2x<4x-19

-6x<-19

-6x(-1)<-19(-1)

6x>19

x>19/6

x>3 1/6

just change all the inequality signs to a x is greater or equal to 3 1/6 and on the rest change to and < and equal sign only put > or equal sign on the last three lines

8 0

2 years ago

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form ModifyingBelow lim With h

Gre4nikov [31]

Answer:

$\dfrac{1}{2\sqrt{x}}$

Step-by-step explanation:

$f(x) = \sqrt{x} = x^{\frac{1}{2}}$

$f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}$

We use binomial expansion for $(x+h)^{\frac{1}{2}}$

This can be rewritten as

$[x(1+\dfrac{h}{x})]^{\frac{1}{2}}$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}$

From the expansion

$(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots$

Setting $x=\dfrac{h}{x}$ and $n=\frac{1}{2}$ ,

$(1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots$

$=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots$

Multiplying by $x^{\frac{1}{2}}$ ,

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots$

The limit of this as $h\to 0$ is

$\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}}$ (since all the other terms involve $h$ and vanish to 0.)

8 0

2 years ago