Please hurry with this do now question

sleet_krkn [62]

<span>1 - 2.25x</span>⁸ =
1² - (1.5x⁴)² =
(1 + 1.5x⁴)(1 - 1.5x⁴)

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bp%7D%7B%20-%208%7D%20%20%2B%209%20%3E%2013" id="TexFormula1" title=" \frac{p}{ -

Gnom [1K]

is the answer open circle?

7 0

3 years ago

Read 2 more answers

In an ore, 9.8% of its total weight is metal. How many pounds of metal are in 1,950 lb of ore?

Fudgin [204]

Answer

Find out the how many pounds of metal are in 1,950 lb of ore .

To proof

let us assume that the pounds of metal are in 1,950 lb of ore be x .

As given

In an ore, 9.8% of its total weight is metal.

ore weight = 1,950 lb

9.8% is written in the decimal form

$= \frac{9.8}{100}$

= 0.098

Than the equation becomes

x = 0.098 × 1950

x = 191.1 pounds

Therefore the 191.1 pounds of metal are in 1,950 lb of ore .

Hence proved

5 0

3 years ago

a shopkeeper sold a certain number ( a two-digit number)of toys all priced at a certain value (also a two-digit number when expr

Makovka662 [10]

The answer is 91 toys sold, make the number ab where a is the 10th digit and b is the first digit. The value is 10a + b that can expressed as 10 (3) + 4 = 34

Let the price of each item: xy

10x + y

He accidentally reversed the digits to: 10b + a toys sold at 10y + x rupees per toy. To get use the formula, he sold 10a + b toys but thought he sold 10b + a toys. The number of toys that he thought he left over was 72 items more than the actual amount of toys left over. So he sold 72 more toys than he thought:

10a + b =10b + a +72

9a = 9b + 72

a = b + 8

The only numbers that could work are a = 9 and b = 1 since a and b each have to be 1 digit numbers. He reversed the digits and thought he sold 19 toys. So the actual number of toys sold was 10a + b = 10 (9) + 1 = 91 toys sold. By checking, he sold 91 – 19 = 72 toys more than the amount that he though the sold. As a result, the number of toys he thought he left over was 72 more than the actual amount left over as was stated in the question.

3 0

3 years ago

Name/ Uid:1. In this problem, try to write the equations of the given surface in the specified coordinates.(a) Write an equation

Gemiola [76]

To find:

(a) Equation for the sphere of radius 5 centered at the origin in cylindrical coordinates

(b) Equation for a cylinder of radius 1 centered at the origin and running parallel to the z-axis in spherical coordinates

Solution:

(a) The equation of a sphere with center at (a, b, c) & having a radius 'p' is given in cartesian coordinates as:

$(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=p^{2}$

Here, it is given that the center of the sphere is at origin, i.e., at (0,0,0) & radius of the sphere is 5. That is, here we have,

$a=b=c=0,p=5$

That is, the equation of the sphere in cartesian coordinates is,

$(x-0)^{2}+(y-0)^{2}+(z-0)^{2}=5^{2}$

$\Rightarrow x^{2}+y^{2}+z^{2}=25$

Now, the cylindrical coordinate system is represented by $(r, \theta,z)$

The relation between cartesian and cylindrical coordinates is given by,

$x=rcos\theta,y=rsin\theta,z=z$

$r^{2}=x^{2}+y^{2},tan\theta=\frac{y}{x},z=z$

Thus, the obtained equation of the sphere in cartesian coordinates can be rewritten in cylindrical coordinates as,

$r^{2}+z^{2}=25$

This is the required equation of the given sphere in cylindrical coordinates.

(b) A cylinder is defined by the circle that gives the top and bottom faces or alternatively, the cross section, & it's axis. A cylinder running parallel to the z-axis has an axis that is parallel to the z-axis. The equation of such a cylinder is given by the equation of the circle of cross-section with the assumption that a point in 3 dimension lying on the cylinder has 'x' & 'y' values satisfying the equation of the circle & that 'z' can be any value.

That is, in cartesian coordinates, the equation of a cylinder running parallel to the z-axis having radius 'p' with center at (a, b) is given by,

$(x-a)^{2}+(y-b)^{2}=p^{2}$

Here, it is given that the center is at origin & radius is 1. That is, here, we have, $a=b=0,p=1$ . Then the equation of the cylinder in cartesian coordinates is,

$x^{2}+y^{2}=1$

Now, the spherical coordinate system is represented by $(\rho,\theta,\phi)$

The relation between cartesian and spherical coordinates is given by,

$x=\rho sin\phi cos\theta,y=\rho sin\phi sin\theta, z= \rho cos\phi$

Thus, the equation of the cylinder can be rewritten in spherical coordinates as,

$(\rho sin\phi cos\theta)^{2}+(\rho sin\phi sin\theta)^{2}=1$

$\Rightarrow \rho^{2} sin^{2}\phi cos^{2}\theta+\rho^{2} sin^{2}\phi sin^{2}\theta=1$

$\Rightarrow \rho^{2} sin^{2}\phi (cos^{2}\theta+sin^{2}\theta)=1$

$\Rightarrow \rho^{2} sin^{2}\phi=1$ (As $sin^{2}\theta+cos^{2}\theta=1$ )

Note that $\rho$ represents the distance of a point from the origin, which is always positive. $\phi$ represents the angle made by the line segment joining the point with z-axis. The range of $\phi$ is given as $0\leq \phi\leq \pi$ . We know that in this range the sine function is positive. Thus, we can say that $sin\phi$ is always positive.

Thus, we can square root both sides and only consider the positive root as,

$\Rightarrow \rho sin\phi=1$

This is the required equation of the cylinder in spherical coordinates.

Final answer:

(a) The equation of the given sphere in cylindrical coordinates is $r^{2}+z^{2}=25$

(b) The equation of the given cylinder in spherical coordinates is $\rho sin\phi=1$

7 0

3 years ago