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3 years ago

What is the answer??? How can we solve it?

Mathematics

2 answers:

GenaCL600 [577]3 years ago

4 0

Okay, so it's pretty easy.
Do 11*11=121

times that by 1/2

so 60.5
v=60.5(6)
now times that by 6
v=363

suter [353]3 years ago

3 0

Hello There, This is Apple Ware Answering Assistant!I have an answer!!!Your answer is: V=363Explanation: 1 Simplify {11}^{2}112 to 121121
V=\frac{1}{2}\times 121\times 6V=21×121×6
2 Simplify \frac{1}{2}\times 121\times 621×121×6 to \frac{726}{2}2726
V=\frac{726}{2}V=2726
3 Simplify \frac{726}{2}2726 to 363363
Please rate me with 5 Stars for better answers!!! AppleWare BETA v1.2.4

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Lunna [17]

D = drinks, p = popcorn
17 + 3.50d + 3.00p

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6 0

4 years ago

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 10.4

vlabodo [156]

Answer:

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

Step-by-step explanation:

Bernoulli's equation,

$P_1+\frac12 \rho v^2_1+\rho g h_1= P_2+\frac12 \rho v^2_2+\rho g h_2$

P₁ = P₂= atmospheric presser

$\rho$ = density

$\frac12 \rho v^2_1+\rho g h_1= \frac12 \rho v^2_2+\rho g h_2$ [since P₁ = P₂]

$\Rightarrow\rho (\frac12 v^2_1+ g h_1)= \rho(\frac12 v^2_2+ g h_2)$

$\Rightarrow\frac12 v^2_1+ g h_1= \frac12 v^2_2+ g h_2$

$\Rightarrow\frac12 v^2_2-\frac12 v^2_1=g h_1- g h_2$

$\Rightarrow v^2_2- v^2_1=2g h$ [ $h_1-h_2=h$ ]

Here $v_1\approx 0$

$\Rightarrow v^2_2=2g h$

$\therefore v_2=\sqrt {2gh$

Here g= 9.8 m/s² , h = 10.4 m

The velocity of water that leaves from the hole $v_2$ = $\sqrt {2\times 9.8\times 10.4}$ m/s

=14.28 m/s.

Given, the rate of flow from the leak is $2.53\times 10^{-3}$ $m^3/min$

$=\frac{2.53\times 10^{-3}}{60}$ $m^3/s$

Let the diameter of the hole be d.

Then the cross section area of the hole is $=\pi (\frac d2)^2$

We know that,

The rate of flow = Cross section area × speed

$\Rightarrow \frac{2.53\times 10^{-3}}{60} =\pi (\frac d2)^2\times 14.28$

$\Rightarrow (\frac d2)^2=\frac{2.53\times 10^{-3}}{60\times 14.28\times \pi}$

$\Rightarrow d= 1.94 \times 10^{-3}$

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

4 0

4 years ago

What's 5 5/6 as a improper fraction

yawa3891 [41]

Answer: 35/6
Work: 5*6=30
30+5=35
35/6

7 0

4 years ago

A woman is emptying her aquarium at a steady rate with a small pump. The water pumped to a 12-in.-diameter cylindrical bucket, a

Temka [501]

Answer:

Therefore the rate at which water level is dropping is $\frac{11}{21}$ in per minute.

Step-by-step explanation:

Given that,

The diameter of cylindrical bucket = 12 in.

Depth is increasing at the rate of = 4.0 in per minutes.

i.e $\frac{dh_1}{dt}=4$

$h_1$ is depth of the bucket.

The volume of the bucket is V = $\pi r^2 h$

$=\pi \times 6^2\itimes h_1$

$\therefore V=36\pi h_1$

Differentiating with respect yo t,

$\frac{dV}{dt}=36\pi \frac{dh_1}{dt}$

Putting $\frac{dh_1}{dt}=4$

$\therefore\frac{dV}{dt}=36\pi\times 4$

The rate of volume change of the bucket = The rate of volume change of the aquarium .

Given that,The aquarium measures 24 in × 36 in × 18 in.

When the water pumped out from the aquarium, the depth of the aquarium only changed.

Consider h be height of the aquarium.

The volume of the aquarium is V= ( 24× 36 ×h)

V= 24× 36 ×h

Differentiating with respect to t

$\frac{dV}{dt}=24\times 36 \times \frac{dh}{dt}$

Putting $\frac{dV}{dt}=36\pi\times 4$

$36\pi\times 4= 24\times 36\times \frac{dh}{dt}$

$\Rightarrow \frac{dh}{dt}=\frac{36\pi \times 4}{24\times 36}$

$\Rightarrow \frac{dh}{dt}=\frac{11}{21}$

Therefore the rate at which water level is dropping is $\frac{11}{21}$ in per minute.

7 0

4 years ago

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Oksi-84 [34.3K]

Answer:

Hey there!

First, let's identify the point. We always go x axis first, then y axis, so the point is at (3, 8).

From the graph, we see that it means the tree is 8 feet tall and three years old.

Let me know if this helps :)

5 0

3 years ago