Melissa made a school banner 3 yards long and 11/12 yard wide what is the area of the banner?

UNLI Quiz Active 1 2 3 5 11 TIME What should be done to solve the equation? X+ 14 = 21 Add 14 to both sides of the equation. Sub

sladkih [1.3K]

I need help to and no one gives me help sjsjsj

7 0

3 years ago

Read 2 more answers

an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a

viktelen [127]

Answer:

the rate of change of the water depth when the water depth is 10 ft is; $\mathbf{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at a time t and r = radius of the cone shaped at a time t

Then the similar triangles ΔOCD and ΔOAB is as follows:

$\dfrac{h}{r}= \dfrac{20}{8}$ ( similar triangle property)

$\dfrac{h}{r}= \dfrac{5}{2}$

$\dfrac{h}{r}= 2.5$

h = 2.5r

$r = \dfrac{h}{2.5}$

The volume of the water in the tank is represented by the equation:

$V = \dfrac{1}{3} \pi r^2 h$

$V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h$

$V = \dfrac{1}{18.75} \pi \ h^3$

The rate of change of the water depth is :

$\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

Since the water is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

$\dfrac{dv}{dt}= - 4 \ ft^3/sec$

Therefore,

$-4 = \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

the rate of change of the water at depth h = 10 ft is:

$-4 = \dfrac{ 100 \ \pi }{6.25}\ \dfrac{dh}{dt}$

$100 \pi \dfrac{dh}{dt} = -4 \times 6.25$

$100 \pi \dfrac{dh}{dt} = -25$

$\dfrac{dh}{dt} = \dfrac{-25}{100 \pi}$

Thus, the rate of change of the water depth when the water depth is 10 ft is; $\mathtt{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

4 0

4 years ago

200 kg of brass is melted down and cast into ornamental frogs, each weight 3/20kg. How many frogs are made?

Savatey [412]

Answer:

1,333 frogs

Step-by-step explanation:

okay so we know that mass remains conserved no matter where you are.

same here ^^

if the total mass of brass is 200 kg

the total mass of the new frogs formed from it when put together will be the same :)

and if there were n such frogs formed

we have,

total mass = n × mass of each frog

200 = n × 3/ 20

n = 4000/ 3

so it becomes something like 1,333.33

that is nearly 1,333 frogs can be made out of 200 kg of brass each weighing 3/ 20kg (the last ones a bit less to make upto 200kg)

5 0

3 years ago

For mutually exclusive events we can use the ____ Principle of Counting.

leonid [27]

Fundamental principle of counting

5 0

3 years ago

Please answer correctly !!!!!! Will mark brainliest !!!!!!!!!!!!

irinina [24]

Answer:

There were 12 babies and 4 adults

8 0

3 years ago