Whooh
SA=4pir²
take derivitive
dSA/dt=8pir dr/dt
and do the volume as well
V=(4/3)pir³
dV/dt=4pir² dr/dt
we need to solve for dV/dt
to do taht we need dr/dt
so
dSA/dt=8pir dr/dt
given dSA/dt=3pi cm/sec
r=2
3pi=8pi2 dr/dt
3=16 dr/dt
3/16=dr/dt
now do the volume
dV/dt=4pir² dr/dt
r=2
dr/dt=3/16
dV/dt=4pi2² (3/16)
dV/dt=16pi(3/16)
dV/dt=3pi
nice
the volume of the sphere is decreasing at 3pi cm/sec as well
I will try my best to assist you with your question ,
First we must divide the amount by 2
334/2 = 167
Now we add 16
167 + 16=183
So, the girl spent $183
Wow, she is spoiled!
Hope I helped!
Good Luck!
Answer:
24 munchkins.
Step-by-step explanation:
Let C be the number of chocolate and D be number of glazed donut holes in the original box.
We are told if Jacob ate 2 chocolate munchkins, then 1/11 of the remaining Munchkins would be chocolate. We can represent this information as:
We are also told if he instead added 4 glazed Munchkins to the original box, 1/7 of the Munchkins would be chocolate. We can represent this information as:
Upon substituting C's value from equation (2) in equation (1) we will get,
Let us have a common denominator on right side of equation.
Multiplying both sides of our equation by 7, we will get,
Multiplying both sides of our equation by 11, we will get,
Therefore, the total number of Munchkins in original box is 24.
Answer:
6(6x^2-4x-3)
Step-by-step explanation:
pull out the gcf which is 6 and it is complete because it is impossible to factor 6x^2-4x-3
First, find 25% of 49.95. Turn 25% into a decimal by dividing by 100.
25% ⇒ 0.25
Now, multiply 0.25 by 49.95.
0.25 × 49.95 = 12.46
25% of 49.95 is 12.46, so subtract 12.46 from 49.95.
49.95 - 12.46 = 37.49
<h2>Answer:</h2>
<u>The discounted price of the jacket is </u><u>$37.49</u><u>.</u>
Now find 8% of 37.49.
8% ⇒ 0.08
Multiply 0.08 by 37.49.
0.08 × 37.49 = 2.99
<h2>Answer:</h2>
<u>The tax is </u><u>$2.99</u><u>.</u>
To find the full price, add 37.49 and 2.99 together.
37.49 + 2.99 = 40.48
<h2>Answer:</h2>
<u>The final cost is </u><u>$40.48</u><u>.</u>
