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2 years ago

Can you match the teacher's comments to the definitions?

1 answer:

3 0

Answer:

Daniel =Were you thinking of perpendicular lines?

Oris= It seems like you've got the basic idea, but your definition could be more precise.

Kaoris= Yes well done...

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blagie [28]

The answer to this question is the second choice
hope this helps:)

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3 years ago

hammer [34]

Answer:

B&C are the answers.

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3 years ago

Aleks04 [339]

Using I=PRT/100
replacing the given information;
(p+prt)=P×r×t

⇒p+prt=P×rt
P=(p+prt)/(rt)
Principal(P)= (p/rt)+p
=p((1/rt)+1)

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3 years ago

Simora [160]

Consider expanding the right hand side as

$y=\sqrt[3]{\dfrac{x(x-2)}{x^2+1}}=x^{1/3}(x-2)^{1/3}(x^2+1)^{-1/3}$

Then taking the logarithm of both sides and applying some properties of the logarithm, you have

$\ln y=\dfrac13\ln x+\dfrac13\ln(x-2)-\dfrac13\ln(x^2+1)$

Now differentiate both sides with respect to $x$ :

$\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1{3x}+\dfrac1{3(x-2)}-\dfrac{2x}{3(x^2+1)}=\dfrac23\dfrac{x^2+x-1}{x(x-2)(x^2+1)}$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac23\dfrac{x^2+x-1}{x(x-2)(x^2+1)}y$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac23\dfrac{x^2+x-1}{x(x-2)(x^2+1)}\sqrt[3]{\dfrac{x(x-2)}{x^2+1}}$

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3 years ago

aliya0001 [1]

Answer:

that he lines are not going strait

Step-by-step explanation:

3 0

3 years ago