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3 years ago

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Challenge:A swimming pool has two inflows and one oulow. Determine their maximal flowspeed in swimming pools per hour if you kno

w the following:Inially, the pool was full.Then, the oulow and one of the inflows were open for minutes.Aerwards, the pool was full.Next, both inflows and the oulow were open for minutes.Aerwards, of the pool was filled with water.Finally, the operator noced that the oulow was open the whole me so he could not fill upthe pool as fast as possible. He closed the oulow and in the next minutes, the pool became full.

1 answer:

insens350 [35]3 years ago

6 0

Answer:

Swimming is a lot of fun, but drowning is a real danger. Even kids who know how to swim can drown, so let's find out how to stay safe in the water.

Why Is It Important to Be Safe in the

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All 6 members of the Valastro family have decided to

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Answer:

5

Step-by-step explanation:

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2 years ago

Which graph shows the equation c = 10+ 3t, where c is the total cost of going to the carnival and t is the number of $3 tickets

lubasha [3.4K]

we have

$c = 10+ 3t$

where

c---------> is the total cost of going to the carnival

t---------> is the number of $\$3$ tickets purchased

we know that

<u>the domain of the function is the interva</u>l---------------> [0,∞)

Because the number of tickets can not be negative

<u>The range of the function is the interval</u>-------------> [10,∞)

Because the total cost can not be negative

using a graph tool

see the attached figure

The answer in the attached figure

7 0

3 years ago

Read 2 more answers

The measure of angle A is 15°, and the length of side

dangina [55]

Answer:

Part 1) $AB=30.9\ units$

Part 2) $AC=29.9\ units$

Step-by-step explanation:

The picture of the question in the attached figure

Part 1

Find the length side AB

we know that

$sin(A)=\frac{BC}{AB}$ ----> by SOH (opposite side divided by the hypotenuse)

substitute the given values

$sin(15^o)=\frac{8}{AB}$

solve for AB

$AB=\frac{8}{sin(15^o)}=30.9\ units$

Part 2

Find the length side AC

we know that

$tan(A)=\frac{BC}{AC}$ ----> by TOA (opposite side divided by the adjacent side)

substitute the given values

$tan(15^o)=\frac{8}{AC}$

solve for AC

$AC=\frac{8}{tan(15^o)}=29.9\ units$

8 0

3 years ago

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

How do you solve this problem?

Yuri [45]

-m-4=-13
-m=-13+4=-9
m=9
☺☺☺☺

8 0

3 years ago

Read 2 more answers