HELP ME PLEASE!!!!!! ASAP!!!!!!!

4. Write the equation in point-slope form. (-6, -2) and m = 3

yKpoI14uk [10]

Y+2=3(x+6) is your answer.

5 0

2 years ago

A punch glass is in the shape of a hemisphere with a radius of 5 cm. If the punch is being poured into the glass so that the cha

Galina-37 [17]

Answer:

28.27 cm/s

Step-by-step explanation:

Though Process:

The punch glass (call it bowl to have a shape in mind) is in the shape of a hemisphere
the radius $r=5cm$
Punch is being poured into the bowl
The height at which the punch is increasing in the bowl is $\frac{dh}{dt} = 1.5$
the exposed area is a circle, (since the bowl is a hemisphere)
the radius of this circle can be written as $'a'$

what is being asked is the rate of change of the exposed area when the height $h = 2 cm$
the rate of change of exposed area can be written as $\frac{dA}{dt}$ .
since the exposed area is changing with respect to the height of punch. We can use the chain rule: $\frac{dA}{dt} = \frac{dA}{dh} . \frac{dh}{dt}$

and since $A = \pi a^2$ the chain rule above can simplified to $\frac{da}{dt} = \frac{da}{dh} . \frac{dh}{dt}$ -- we can call this Eq(1)

Solution:

the area of the exposed circle is

$A =\pi a^2$

the rate of change of this area can be, (using chain rule)

$\frac{dA}{dt} = 2 \pi a \frac{da}{dt}$ we can call this Eq(2)

what we are really concerned about is how $a$ changes as the punch is being poured into the bowl i.e $\frac{da}{dh}$

So we need another formula: Using the property of hemispheres and pythagoras theorem, we can use:

$r = \frac{a^2 + h^2}{2h}$

and rearrage the formula so that a is the subject:

$a^2 = 2rh - h^2$

now we can derivate a with respect to h to get $\frac{da}{dh}$

$2a \frac{da}{dh} = 2r - 2h$

simplify

$\frac{da}{dh} = \frac{r-h}{a}$

we can put this in Eq(1) in place of $\frac{da}{dh}$

$\frac{da}{dt} = \frac{r-h}{a} . \frac{dh}{dt}$

and since we know $\frac{dh}{dt} = 1.5$

$\frac{da}{dt} = \frac{(r-h)(1.5)}{a}$

and now we use substitute this $\frac{da}{dt}$ . in Eq(2)

$\frac{dA}{dt} = 2 \pi a \frac{(r-h)(1.5)}{a}$

simplify,

$\frac{dA}{dt} = 3 \pi (r-h)$

This is the rate of change of area, this is being asked in the quesiton!

Finally, we can put our known values:

$r = 5cm$

$h = 2cm$ from the question

$\frac{dA}{dt} = 3 \pi (5-2)$

$\frac{dA}{dt} = 9 \pi cm/s// or//\frac{dA}{dt} = 28.27 cm/s$

5 0

2 years ago

Identify the domain and range of the function below. <br><br> Domain<br><br> Range

scoundrel [369]

Answer:sry

Step-by-step explanation:

6 0

2 years ago

Dailene has 30 CD's and buys two new ones each week. Renae has 18 CD's and buys 4 new ones each week. after now many weeks will

Naddika [18.5K]

6 Weeks is the awnser

3 0

2 years ago

Can somebody please help me? I don't understand this!

Xelga [282]

Use the base of the triangle as the diameter, which in this case is 16, divide it by 2, which is 8, and then square it, which is 64, and multiply by Pi (64 times Pi is 200.96) so you can find the area of the whole circle which is 200.96ft^2. Since this is an equilateral triangle, the height might be 16, so using the area of a triangle, A=BH times 1/2, We multiply 16 by 16 which is 256. 256 times 1/2 is 128. So the area of the triangle will be 128ft^2. We then subtract our areas, 200.96-128 which is 72.96. Then therefore the area of the circle will be 72.96ft^2. I may be wrong, but this is what I think.

5 0

3 years ago