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3 years ago

What is 3.78 divided by 4.2?

Mathematics

2 answers:

mote1985 [20]3 years ago

8 0

I used a calculator.
3.78/ 4.2 = 0.9
Answer: 0.9

Anna [14]3 years ago

5 0

The answer it this problem is 0.9

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Which of the following pairs of terms are like terms?

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If the two beside the p in d is an exponent I believe that would be correct. if not sorry

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2 years ago

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oksano4ka [1.4K]

Answer:

9 cm.

Step-by-step explanation:

Let the height of the pyramid be h cm, then the height of the cuboid is (15 - h) cm.

Volume of the pyramid:

= 1/3 * h * s^2 where s is the length of a side of the square base.

= hs^2/3 cm^3

Volume of the cuboid:

= s^2(15 - h).

So we have:

hs^2/ 3 = 300.......................(1)

s^2(15 - h) = 600..................(2)

From equation (1) :

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Now substitute for s^2 in equation (2) :

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2 years ago

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Step-by-step explanation:

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2 years ago

What is 2 times 5+3 times 2+4

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Step-by-step explanation:

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2 years ago

Read 2 more answers

A tank has the shape of a surface generated by revolving the parabolic segment y = x2 for 0 ≤ x ≤ 3 about the y-axis (measuremen

Darina [25.2K]

Answer:

$100\pi\int\limits^9_0 {(\sqrt y)^2(14-y)} \, dy$ ft-lbs.

Step-by-step explanation:

Given:

The shape of the tank is obtained by revolving $y=x^2$ about y axis in the interval $0\leq x\leq 3$ .

Density of the fluid in the tank, $D=100\ lbs/ft^3$

Let the initial height of the fluid be 'y' feet from the bottom.

The bottom of the tank is, $y(0)=0^2=0$

Now, the height has to be raised to a height 5 feet above the top of the tank.

The height of top of the tank is obtained by plugging in $x=3$ in the parabolic equation . This gives,

$H=3^2=9\ ft$

So, the height of top of tank is, $y(3)=H=9\ ft$

Now, 5 ft above 'H' means $H+5=9+5=14$

Therefore, the increase in height of the top surface of the fluid in the tank is given as:

$\Delta y=(14-y)$ ft

Now, area of cross section of the tank is given as:

$A(y)=\pi r^2\\r\to radius\ of\ the\ cross\ section$

Radius is the distance of a point on the parabola from the y axis. This is nothing but the x-coordinate of the point.

We have, $y=x^2$

So, $x=\sqrt y$

Therefore, radius, $r=\sqrt y$

Now, area of cross section is, $A(y)=\pi (\sqrt y)^2$

Work done in pumping the contents to 5 feet above is given as:

$W=D\int\limits^{y(3)}_{y(0)} {A(y)(\Delta y)} \, dy$

Plug in all the values. This gives,

$W=100\int\limits^9_0 {\pi (\sqrt y)^2(14-y)} \, dy\\\\W=100\pi\int\limits^9_0 { (\sqrt y)^2(14-y)} \, dy\textrm{ ft-lbs}$

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2 years ago