Can you please help me fast

Mathematics

1 answer:

DaniilM [7]3 years ago

4 0

Find the values of.....?

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I need help with this

SSSSS [86.1K]

Answer:13 thank me later

8 0

3 years ago

Which is the right answer?????

Inga [223]

Answer: A

Step-by-step explanation: put it into point slope-form

y-ysub1=m(x-xsub1)

m=slope=(-2-3)/(2-(-1))=-5/3

ysub1 is 3 and xsub1 is -1

so substitute: y-3=-5/3(x-(-1))

y-3=-5/2(x+1)

6 0

3 years ago

Calculate the following limit:

aleksklad [387]

$\lim_{x\to\infty}\dfrac{\sqrt x}{\sqrt{x+\sqrt{x+\sqrt x}}}=\\
\lim_{x\to\infty}\dfrac{\dfrac{\sqrt x}{\sqrt x}}{\dfrac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt x}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{\dfrac{x+\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{\sqrt{x^2}}}}=\\$
$\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{x+\sqrt x}{x^2}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{\sqrt x}{\sqrt{x^4}}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{x}{x^4}}}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{1}{x^3}}}}}=\\
=\dfrac{1}{\sqrt{1+\sqrt{0+\sqrt{0}}}}=\\$
$=\dfrac{1}{\sqrt{1+0}}=\\
=\dfrac{1}{\sqrt{1}}=\\
=\dfrac{1}{1}=\\
1
$

8 0

3 years ago

Help will mark Brainliest graph is then show me a pic pls

gregori [183]

Answer:

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