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3 years ago

13

Can anybody please help me with question 1 please...I would really appreciate it so much

2 answers:

In-s [12.5K]3 years ago

5 0

Answer:

The total time the water balloon is in the air is 8 seconds. The balloon reaches the maximum height of 25 fee after 3 seconds. The water balloon has a height of 24 feet at 2 seconds and a half.

Step-by-step explanation:

exis [7]3 years ago

5 0

Answer:

I was saying the same thing

Step-by-step explanation:

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cylinder shaped can needs to be constructed to hold 200 cubic centimeters of soup. The material for the sides of the can costs 0

Dafna11 [192]

Answer:

Radius=2.09 cm

Height,h=14.57 cm

Step-by-step explanation:

We are given that

Volume of cylinderical shaped can=200 cubic cm.

Cost of sides of can=0.02 cents per square cm

Cost of top and bottom of the can =0.07 cents per square cm

Curved surface area of cylinder= $2\pi rh$

Area of circular base=Area of circular top= $\pi r^2$

Total cost,C(r)= $0.02\times 2\pi rh+2\pi r^2\times 0.07$

Volume of cylinder, $V=\pi r^2 h$

$200=\pi r^2 h$

$h=\frac{200}{\pi r^2}$

Substitute the value of h

$C(r)=0.02\times 2\pi r\times \frac{200}{\pi r^2}+2\pi r^2\times 0.07$

$C(r)=\frac{8}{r}+0.14\pi r^2$

Differentiate w.r.t r

$C'(r)=-\frac{8}{r^2}+0.28\pi r$

$C'(r)=0$

$-\frac{8}{r^2}+0.28\pi r=0$

$0.28\pi r=\frac{8}{r^2}$

$r^3=\frac{8}{0.28\pi}=9.095$

$r=(9.095)^{\frac{1}{3}}=2.09$

Again, differentiate w.r.t r

$C''(r)=\frac{16}{r^3}+0.28\pi$

Substitute the value of r

$C''(2.09)=\frac{16}{(2.09)^3}+0.28\pi=2.63>0$

Therefore,the product cost is minimum at r=2.09

h= $\frac{200}{\pi (2.09)^2}=14.57$

Radius of can,r=2.09 cm

Height of cone,h=14.57 cm

4 0

3 years ago