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LenKa [72]
3 years ago
15

Question 7 il pont) (2 +377)(2 – 377)

Mathematics
1 answer:
Elden [556K]3 years ago
3 0

Answer:

142125

Step-by-step explanation:

Since we given two brackets to solve we will have to take the step by step procedure

So let's solve the question

(2+377)(2-377)

Let's start with the first bracket

(2+377)

It will give us 379

So let's solve the second bracket

(2-377)

It will give us -375

So back to the question

(379)(-375)

142125

So the final answer is 142125

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1/2 x + 3 = 15 what is the x
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Answer:

x=24

Step-by-step explanation:

x/2+3=15

-3 both sides

x/2=15−3

15-3=12

x/2=12

12*2=24

x=24

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3 years ago
Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 4cm and a height of 16cm, at the ra
bearhunter [10]

Answer:

\frac{dh}{dt}=-\frac{1}{2\pi}cm/min

Step-by-step explanation:

From similar triangles, see diagram in attachment

\frac{r}{4}=\frac{h}{16}


We solve for r to obtain,


r=\frac{h}{4}


The formula for calculating the volume of a cone is

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


We substitute the value of r=\frac{h}{4} to obtain,


V=\frac{1}{3}\pi (\frac{h}{4})^2h


This implies that,

V=\frac{1}{48}\pi h^3


We now differentiate both sides with respect to t to get,

\frac{dV}{dt}=\frac{\pi}{16}h^2 \frac{dh}{dt}


We were given that water is drained out of the tank at a rate of 2cm^3/min


This implies that \frac{dV}{dt}=-2cm^3/min.


Since we want to determine the rate at which the depth of the water is changing at the instance when the water in the tank is 8cm deep, it means h=8cm.


We substitute this values to obtain,


-2=\frac{\pi}{16}(8)^2 \frac{dh}{dt}


\Rightarrow -2=4\pi \frac{dh}{dt}


\Rightarrow -1=2\pi \frac{dh}{dt}


\frac{dh}{dt}=-\frac{1}{2\pi}






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