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3 years ago

Help me .........................................................

Mathematics

2 answers:

qwelly [4]3 years ago

8 0

Answer:

Step-by-step explanation:

I don't know what your teacher thinks about solving this in terms of slope/ y intercept. I'll it that way. It makes things much easier to graph which I'll also include.

y - 2 = -1/2(x + 3) Remove the brackets.

y - 2 = -1/2x - 1.5 Add 2 to both sides.

y = -1/2x - 1.5 + 2

y = -1/2x + 1/2

You can make a small table if you like

x y

0 1/2

1 0

2 -1/2

You should check the last entry

y = -1/2 * 2 + 1/2

y = -1 + 1/2

y = -1/2

Jet001 [13]3 years ago

4 0

Answer:

the answer is y= -1/2x+1/2

Step-by-step explanation:

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When the area in square units of an expanding circle is increasing twice as fast as its radius in linear units, the radius is...

enot [183]

Answer:

c. $\frac{1}{\pi}$

Step-by-step explanation:

We have been given that the area in square units of an expanding circle is increasing twice as fast as its radius in linear units

We will use derivatives to solve our given problem.

We know that area (A) of a circle is equal to $A=\pi r^2$ .

Let us find derivative of area function with respect to time.

$\frac{dA}{dt}=\frac{d}{dt}(\pi r^2)$

Bring out constant:

$\frac{dA}{dt}=\pi \frac{d}{dt}(r^2)$

Using power rule and chain rule, we will get:

$\frac{dA}{dt}=\pi(2r)* \frac{dr}{dt}$

$\frac{dA}{dt}=2\pi r*\frac{dr}{dt}$ Here $\frac{dr}{dt}$ represents change is radius with respect to time.

We have been given that area of an expanding circle is increasing twice as fast as its radius in linear units. We can represent this information in an equation as:

$\frac{dA}{dt}=2\frac{dr}{dt}$