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myrzilka [38]
3 years ago
8

Find the cuberoot of 512please answer fast thank you​

Mathematics
2 answers:
Stolb23 [73]3 years ago
6 0

Answer:

8 is the cube root of 512

Step-by-step explanation:

=》

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

=》

\sqrt[3]{512}  = 2 \times 2 \times 2

=》

\sqrt[3]{512}  = 8

RoseWind [281]3 years ago
3 0

Answer:

8

Step-by-step explanation:

\sqrt[3]{512}  =  \sqrt[3]{ {8}^{3} } = 8 \\

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Svetach [21]
When it comes to dividing decimals, bear in mind that, first off you'd want to do away with the decimal point.

now, "you add as many zeros to the other term, as it has decimals", what the heck does that mean?

for example 2.135 ÷ 5

notice, the 2.135 has three decimals, namely three numbers to the right of the decimal point.

since 2.135 has three decimals, then what we do, we add three zeros to 5, AND do away with the decimal point, thus our division is really

2135 ÷ 5000, which is 0.427

now, if we had 2.135 ÷ 5.79

what we do is, since 2.135 has three decimals, we add three zeros to 5.79, and since 5.79 has two decimals, we add two zeros to 2.135, then our division is really

213500 ÷ 579000  <-- notice, no dot on both

which is about 0.3687392

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4 years ago
A water tank is in the shape of a cone.Its diameter is 50 meter and slant edge is also 50 meter.How much water it can store In i
Aneli [31]
To get the most accurate answer possible, we're going to have to go into some unsightly calculation, but bear with me here:

Assessing the situation:

Let's get a feel for the shape of the problem here: what step should we be aiming to get to by the end? We want to find out how long it will take, in minutes, for the tank to drain completely, given a drainage rate of 400 L/s. Let's name a few key variables we'll need to keep track of here:

V - the storage volume of our tank (in liters)
t - the amount of time it will take for the tank to drain (in minutes)

We're about ready to set up an expression using those variables, but first, we should address a subtlety: the question provides us with the drainage rate in liters per second. We want the answer expressed in liters per minute, so we'll have to make that conversion beforehand. Since one second is 1/60 of a minute, a drainage rate of 400 L/s becomes 400 · 60 = 24,000 L/min.

From here, we can set up our expression. We want to find out when the tank is completely drained - when the water volume is equal to 0. If we assume that it starts full with a water volume of V L, and we know that 24,000 L is drained - or subtracted - from that volume every minute, we can model our problem with the equation

V-24000t=0

To isolate t, we can take the following steps:

V-24000t=0\\ V=24000t\\ \frac{V}{24000}=t

So, all we need to do now to find t is find V. As it turns out, this is a pretty tall order. Let's begin:

Solving for V:

About units: all of our measurements for the cone-shaped tank have been provided for us in meters, which means that our calculations will produce a value for the volume in cubic meters. This is a problem, since our drainage rate is given to us in liters per second. To account for this, we should find the conversion rate between cubic meters and liters so we can use it to convert at the end.

It turns out that 1 cubic meter is equal to 1000 liters, which means that we'll need to multiply our result by 1000 to switch them to the correct units.

Down to business: We begin with the formula for the area of a cone,

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

which is to say, 1/3 multiplied by the area of the circular base and the height of the cone. We don't know h yet, but we are given the diameter of the base: 50 m. To find the radius r, we divide that diameter in half to obtain r = 50/2 = 25 m. All that's left now is to find the height.

To find that, we'll use another piece of information we've been given: a slant edge of 50 m. Together with the height and the radius of the cone, we have a right triangle, with the slant edge as the hypotenuse and the height and radius as legs. Since we've been given the slant edge (50 m) and the radius (25 m), we can use the Pythagorean Theorem to solve for the height h:

h^2+25^2=50^2\\ h^2+625=2500\\ h^2=1875\\ h=\sqrt{1875}=\sqrt{625\cdot3}=25\sqrt{3}

With h=25\sqrt{3} and r=25, we're ready to solve for V:

V= \frac{1}{3} \pi(25)^2\cdot25\sqrt{3}\\ V= \frac{1}{3} \pi\cdot625\cdot25\sqrt{3}\\ V= \frac{1}{3} \pi\cdot15625\sqrt{3}\\\\ V= \frac{15625\sqrt{3}\pi}{3}

This gives us our volume in cubic meters. To convert it to liters, we multiply this monstrosity by 1000 to obtain:

\frac{15625\sqrt{3}\pi}{3}\cdot1000= \frac{15625000\sqrt{3}\pi}{3}

We're almost there.

Bringing it home:

Remember that formula for t we derived at the beginning? Let's revisit that. The number of minutes t that it will take for this tank to drain completely is:

t= \frac{V}{24000}

We have our V now, so let's do this:

t= \frac{\frac{15625000\sqrt{3}\pi}{3}}{24000} \\ t= \frac{15625000\sqrt{3}\pi}{3}\cdot \frac{1}{24000} \\ t=\frac{15625000\sqrt{3}\pi}{3\cdot24000}\\ t=\frac{15625\sqrt{3}\pi}{3\cdot24}\\ t=\frac{15625\sqrt{3}\pi}{72}\\ t\approx1180.86

So, it will take approximately 1180.86 minutes to completely drain the tank, which can hold approximately V= \frac{15625000\sqrt{3}\pi}{3}\approx 28340615.06 L of fluid.
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Costs of Attendance
erastovalidia [21]

Answer:

y = 3,224x + 750

Step-by-step explanation:

Assuming the scholarship and grants are applied using the slope intercept form y=mx+b where m is the slope in this case the cost per year attending x the number of years attended and b the y intercept which in this case is the one time fee subtracting 13,774 by 10,550 gives us 3,224.

Since the one time fee is only in place once that would make it the y intercept therefore y=3,224x+750 is the correct answer

8 0
3 years ago
What percent of 75.5 is 81.54
lutik1710 [3]
If you would like to know what percent of 75.5 is 81.54, you can calculate this using the following steps:

x% of 75.5 is 81.54
x% * 75.5 = 81.54
x/100 * 75.5 = 81.54
x = 81.54 * 100 / 75.5
x = 108%

Result: 108% of 75.5 is 81.54.
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