All categories

1 year ago

Factor 125x3 + 343y3

Mathematics

1 answer:

Annette [7]1 year ago

5 0

Answer:

(5x + 7y)(25x² - 35xy + 49y²)

Step-by-step explanation:

125x³ + 343y³ ← is a sum of cubes and factors in general as

a³ + b³ = (a + b)(a² - ab + b² )

125x³ = (5x)³ ⇒ a = 5x

343y³ = (7y)³ ⇒ b = 7y

125x³ + 343y³

= (5x + 7y)((5x)² - (5x × 7y) + (7y)²)

= (5x + 7y)(25x² - 35xy + 49y²) ← in factored form

You might be interested in

Annie is creating a stencil for her artwork using a coordinate plane. The beginning of the left edge of the stencil falls at (2,

taurus [48]

Answer:

(A)(12, 9)

Step-by-step explanation:

Given:

The beginning of the left edge of the stencil falls at (2, −1).

A point, say Q on the stencil is at (4, 1).

Point Q divides the stencil into the ratio 1:4.

We are required to find the end of the stencil.

Mathematically, Point Q divides the stencil internally in the ratio 1:4.

For internal division of a line with beginning point $(x_1,y_1)$ and end point $(x_2,y_2)$ in the ratio m:n, we use the formula

$Q(x,y)=(\dfrac{mx_2+nx_1}{m+n} ,\dfrac{my_2+ny_1}{m+n} )$

$(x_1,y_1)=(2, -1)$ , $(x_2,y_2)=?$ , Q(x,y)=(4,1), m:n=1:4

Therefore:

$(4,1)=(\dfrac{1x_2+4*2}{1+4} ,\dfrac{1y_2+4*-1}{1+4} )\\(4,1)=(\dfrac{x_2+8}{5} ,\dfrac{y_2-4}{5} )\\$Therefore:\\\dfrac{x_2+8}{5}=4\\x_2+8=4X5\\x_2=20-8=12\\$Similarly\\\dfrac{y_2-4}{5}=1\\y_2-4=5\\y_2=4+5=9\\(x_2,y_2)=(12,9)$

The correct option is A.

6 0

1 year ago

Read 2 more answers

Mr. Singh needed to understand the trend of his students’ latest test scores. Find the mean and median for each set of data, and

marishachu [46]

Answer:

The Median should be 89.5

And the mean should be 69.75

I hope that helped

9 0

2 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

1 year ago

in the figure p is the incenter of isosceles rst what type of triangle is rpt? Thank you in advance .

snow_lady [41]

Answer:

$\Delta \text{RPT is an isosceles triangle}$

Step-by-step explanation:

As in the triangle RST,

$RS=ST\ \ \ \Rightarrow m\angle SRT=m\angle STR$

Because if in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another.

Incenter is obtained by the Intersection of a triangle's three angle bisectors and it is equidistant from the three sides of the triangle.

So, RP and TP are the angle bisectors of ∠R and ∠T respectively.

$\Rightarrow \dfrac{m\angle SRT}{2}=\dfrac{m\angle STR}{2}$

$\Rightarrow m\angle PRT=m\angle PTR$

$\Rightarrow PR=PT$

$\Rightarrow \Delta \text{RPT is an isosceles triangle}$

7 0

2 years ago

Read 2 more answers

The Greens want to put an addition on their house 18 months from now. They will need to save $10,620 in order to achieve this go

BARSIC [14]

The easiest of the two options to prove is option A.

To see whether or not option A. is true, we need to find out how much the Greens have been saving every month. We know that the Greens have saved $6120 in the last year, or twelve months. To find out how much they have been saving each month, we need to divide the amount they have saved by the number of months they have been saving.

$6120/12 months

This turns out to be $510 each month.

Now, to find out whether or not option 11
A. is true, we need to find out the amount of money that the Greens would save in eighteen months.

$510 * 18 = $9180

Now that we have the amount of money that the Greens would save over eighteen months, we need to see if an additional month of saving would let them reach $10,620.

$9180 + $510 = $9690

Now, we know that option A is not true because the Greens would not have enough money after only one extra month of saving. Because option A is not true, option B must be true.

5 0

2 years ago