Find the volume of the triangular pyramid shown below

Please help me <br> 30 points <br> Show you work

GenaCL600 [577]

Answer:

(a) A translation is a rigid transformation. How does this statement support line s being parallel to line r?

Because ALL of the points of line r move the same distance in the same direction, then the inclination of the line s doesn't change with respect line r.

(b) Write and expression for the slope of line r:

(k-n)/(j-m)

(c) Write and expression for the slope of line s:

(k-n)/(j-m)

(d) Line q is a vertical is a vertical translation of line s 3 units down. P'' is the image of P'. What are the coordinates of P''?

The coordinates of P'' are:

P''=(m,n+ϴ-3)

Step-by-step explanation:

(a) A translation is a rigid transformation. How does this statement support line s being parallel to line r?

Because ALL of the points of line r move the same distance in the same direction, then the inclination of the line s doesn't change with respect line r.

(b) Write and expression for the slope of line r.

Line r goes through the points:

P=(m,n)=(xp,yp)→xp=m, yp=n

Q=(j,k)=(xq,yq)→xq=j, yq=k

Slope of line r is:

mr=(yq-yp)/(xq-xp)

Replacing the coordinates, the expression for the slope of line r is:

mr=(k-n)/(j-m)

(c) Write and expression for the slope of line s.

Line s goes through the points:

P'=(m,n+ϴ)=(xp',yp')→xp'=m, yp'=n+ϴ

Q'=(j,k+ϴ)=(xq',yq')→xq'=j, yq'=k+ϴ

Slope of line s is:

ms=(yq'-yp')/(xq'-xp')

Replacing the coordinates, the expression for the slope of line s is:

ms=[(k+ϴ)-(n+ϴ)]/(j-m)

Eliminating the parentheses in the numerator:

ms=[k+ϴ-n-ϴ]/(j-m)

Simplifyng:

ms=(k-n)/(j-m)

(d) Line q is a vertical is a vertical translation of line s 3 units down. P'' is the image of P'. What are the coordinates of P''?

3 units down, then change only the ordinate y:

P''=(xp',yp'-3)

Replacing xp' and yp':

P''=(m,n+ϴ-3)

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3 years ago

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Find the distance between P1(4,16degrees) and P2(-2,177degrees) on the polar plane.

bazaltina [42]

Polar coordinates give the distance from the origin and the angle from the positive x axis. Cartesian coordinates give the distance from the x and y axes.

You can draw a right triangle with these values. (see attached)
If you know the r value and theta of that triangle below, you can use trig to find x and y.

Let's convert (4, 16°) to Cartesian coordinates.

Note that since our angle is acute, (in Quadrant I) our sine and cosine will both be positive, as you should be able to derive from the unit circle, where cosine is represented as an x value and sine is represented as a y value.

cosine = adjacent / hypotenuse
cosθ = x/r
cos(16°) = x/4
4cos(16°) = x ≈ 3.84504678375

sine = oppsite / hypotenuse
sinθ = y/r
sin(16°) = y/4
4sin(16°) = y ≈ 1.10254942327<span>

So (4, 16°) </span>⇒ (3.84504678375, 1.10254942327).

Let's convert (-2, 177°) to Cartesian coordinates.
Whenever you have a negative radius, that means to put the point opposite where it would have been if it had a positive radius. (see attached)

In that case, we can essentially add 180° to our current 177° to the same effect. That means that (-2, 177°) = (2, 357°).

Note that since our angle is in Quadrant IV, our cosine will be positive, but our sine will be negative. (as derived from the unit circle) We don't have to worry about this since our calculator figures this for us, but you should pay attention to it if you are converting from Cartesian to polar.

cosine = adjacent / hypotenuse
cosθ = x/r
cos(357°) = x/2
2cos(357°) = x ≈ 1.99725906951

sine = opposite / hypotenuse
sinθ = y/r
sin(357°) = y/2
2sin(357°) = y ≈ -0.10467191248

So (-2, 177°) ⇒ (1.99725906951, -0.10467191248).

Now we must use the distance formula with our two points.
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$d\approx\sqrt{(1.99725906951-3.84504678375)^2+(-0.10467191248-1.10254942327)^2}$
$d\approx\sqrt{-1.84778771^2+-1.20722134^2}$
$d\approx\sqrt{3.41431942+1.45738336}$
$d\approx\sqrt{4.87170278}$
$\boxed{d\approx2.20719342}$