<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B75%5D%7B6000%7D%20" id="TexFormula1" title=" \sqrt[75]{6000} " alt=" \sqrt[75]{60

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sleet_krkn [62]

3 years ago

00} " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

PilotLPTM [1.2K]3 years ago

7 0

$\sqrt[75]{6000} =1.12$

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Now answer the question:

Savatey [412]

Answer:

y = 10

Explanation:

a = apples

b = bananas

the a is equal to 1.5, the b is equal to .3

so it's 2a + xb = 6

(2 x 1.5) + (y x 0.3) = 6

3 + (y x 0.3) = 6

y x 0.3 = 3

so the way we get the answer here, we remove the 0.3 from the side that has the number we're looking for. to do that, we put it on the side with the number we know

with addition problems, you just add or subtract from one side and do the same to the other.

with multiplication, we need to divide one side and multiply by the other.

all the dividing we're doing here will be changing the number with the y to be basically not there

but they'd probably have it written as 0.3y = 3

and the way we sometimes show division, is with the / symbol, so to get this, it would be (0.3y/0.3) = 3 once again, parenthesis just to keep it separate and to try and prevent confusion. you remove the 0.3 from one side and put it on the other by doing the same thing to each side

so it's currently (0.3y/0.3) = 3/0.3s
3 divided by 0.3 comes out to 10

so the answer ends up being y = 10

4 0

3 years ago

A total of 816 tickets were sold for the school play. They were either adult tickets or student tickets. There were 66 more stud

Montano1993 [528]

The number of adult tickets sold is the smaller of the two contributors to the sum, so is half the difference of these numbers:
(816 -66)/2 = 375

375 adult tickets were sold.

3 0

4 years ago

Suppose the measure of angle 1 is 127 degrees. Find the measure of the other 3 angles:

alukav5142 [94]

Answer:

Angle 1 is 127°

∠2 is 53°

∠3 is 127°

∠4 is 53°

5 0

4 years ago

HELP ME PLEASE IS IMPORTANT AND ASAP!

zysi [14]

Answer:

maximun (0,5)

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Let D be the smaller cap cut from a solid ball of radius 8 units by a plane 4 units from the center of the sphere. Express the v

natima [27]

Answer:

Step-by-step explanation:

The equation of the sphere, centered a the origin is given by $x^2+y^2+z^2 = 64$ . Then, when z=4, we get

$x^2+y^2= 64-16 = 48$ .

This equation corresponds to a circle of radius $4\sqrt[]{3}$ in the x-y plane

c) We will use the previous analysis to define the limits in cartesian and polar coordinates. At first, we now that x varies from $-4\sqrt[]{3}$ up to $4\sqrt[]{3}$ . This is by taking y =0 and seeing the furthest points of x that lay on the circle. Then, we know that y varies from $-\sqrt[]{48-x^2}$ and $\sqrt[]{48-x^2}$ , this is again because y must lie in the interior of the circle we found. Finally, we know that z goes from 4 up to the sphere, that is , z goes from 4 up to $\sqrt[]{64-x^2-y^2}$

Then, the triple integral that gives us the volume of D in cartesian coordinates is

$\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx$ .

b) Recall that the cylindrical coordinates are given by $x=r\cos \theta, y = r\sin \theta,z = z$ , where r corresponds to the distance of the projection onto the x-y plane to the origin. REcall that $x^2+y^2 = r^2$ . WE will find the new limits for each of the new coordinates. NOte that, we got a previous restriction of a circle, so, since $\theta[\tex] is the angle between the projection to the x-y plane and the x axis, in order for us to cover the whole circle, we need that [tex]\theta$ goes from 0 to $2\pi$ . Also, note that r goes from the origin up to the border of the circle, where r has a value of 4\sqrt[]{3}. Finally, note that Z goes from the plane z=4 up to the sphere itself, where the restriction is \sqrt[]{64-r^2}. So, the following is the integral that gives the wanted volume

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta$ . Recall that the r factor appears because it is the jacobian associated to the change of variable from cartesian coordinates to polar coordinates. This guarantees us that the integral has the same value. (The explanation on how to compute the jacobian is beyond the scope of this answer).

a) For the spherical coordinates, recall that $z = \rho \cos \phi, y = \rho \sin \phi \sin \theta, x = \rho \sin \phi \cos \theta$ . where $\phi$ is the angle of the vector with the z axis, which varies from 0 up to pi. Note that when z=4, that angle is constant over the boundary of the circle we found previously. On that circle. Let us calculate the angle by taking a point on the circle and using the formula of the angle between two vectors. If z=4 and x=0, then y=4 $\sqrt[]{3}$ if we take the positive square root of 48. So, let us calculate the angle between the vector $a=(0,4\sqrt[]{3},4)$ and the vector b =(0,0,1) which corresponds to the unit vector over the z axis. Let us use the following formula

$\cos \phi = \frac{a\cdot b}{||a||||b||} = \frac{(0,4\sqrt[]{3},4)\cdot (0,0,1)}{8}= \frac{1}{2}$

Therefore, over the circle, $\phi = \frac{\pi}{3}$ . Note that rho varies from the plane z=4, up to the sphere, where rho is 8. Since $z = \rho \cos \phi$ , then over the plane we have that $\rho = \frac{4}{\cos \phi}$ Then, the following is the desired integral

$\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{\frac{4}{\cos \phi}}^{8}\rho^2 \sin \phi d\rho d\phi d\theta$ where the new factor is the jacobian for the spherical coordinates.

d ) Let us use the integral in cylindrical coordinates

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta=\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} r (\sqrt[]{64-r^2}-4) dr d\theta=\int_{0}^{2\pi} d \theta \cdot \int_{0}^{4\sqrt[]{3}}r (\sqrt[]{64-r^2}-4)dr= 2\pi \cdot (-2\left.r^{2}\right|_0^{4\sqrt[]{3}})\int_{0}^{4\sqrt[]{3}}r \sqrt[]{64-r^2} dr$

Note that we can split the integral since the inner part does not depend on theta on any way. If we use the substitution $u = 64-r^2$ then $\frac{-du}{2} = r dr$ , then

$=-2\pi \cdot \left.(\frac{1}{3}(64-r^2)^{\frac{3}{2}}+2r^{2})\right|_0^{4\sqrt[]{3}}=\frac{320\pi}{3}$

3 0

3 years ago