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garri49 [273]
3 years ago
14

PLS HELP 50 POINTS

Mathematics
1 answer:
PtichkaEL [24]3 years ago
7 0

Answer:

\text{Volume of prism made by }\Delta XYZ=2304\text{ in}^3  

Step-by-step explanation:

Let us assume that both prisms are similar, so we can use proportions to solve for the volume of prism made by triangle XYZ.

Let us find the proportion between the sides of both triangles.

\frac{\text{Side of triangle XYZ}}{\text{Side of triangle PQR}}=\frac{24}{36}

\frac{\text{Side of triangle XYZ}}{\text{Side of triangle PQR}}=\frac{12*2}{12*3}

\frac{\text{Side of triangle XYZ}}{\text{Side of triangle PQR}}=\frac{2}{3}

Since for volume of triangular prism we multiply base area and height of the prism, this means we will have to multiply the proportion of each side length 3 times to find the proportion of volumes between our both prism.

So we can set proportion for volume of both prisms as:

\frac{\text{Volume of prism made by triangle XYZ}}{\text{Volume of prism made by triangle PQR}}=(\frac{2}{3})^3  

\frac{\text{Volume of prism made by triangle XYZ}}{\text{Volume of prism made by triangle PQR}}=\frac{8}{27}

Upon substituting volume of prism made by triangle PQR we will get,

\frac{\text{Volume of prism made by triangle XYZ}}{7776\text{in}^3}=\frac{8}{27}    

Let us multiply both sides of our equation by 7776 cubic inches.

\frac{\text{Volume of prism made by triangle XYZ}}{7776\text{ in}^3}*7776\text{ in}^3=\frac{8}{27}*7776\text{ in}^3

\text{Volume of prism made by triangle XYZ}=8*288\text{ in}^3

\text{Volume of prism made by triangle XYZ}=2304\text{ in}^3  

Therefore, the volume of prism made by triangle XYZ is 2304 cubic inches.

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Image point B'(4, -8) was transformed using the translation (x-2, y + 3). What were the coordinates of B?
DaniilM [7]

Answer:

B(6 , -11)

Step-by-step explanation:

(x-2 , y+3)  = (4 , -8)

Compare the x & y co ordinates

x - 2 = 4   ; y + 3 = -8

x = 4 +2  ; y = -8 - 3

x = 6      ; y = -11

B(6 , -11)

8 0
3 years ago
The plane x + y + z = 12 intersects paraboloid z = x^2 + y^2 in an ellipse.(a) Find the highest and the lowest points on the ell
emmasim [6.3K]

Answer:

a)

Highest (-3,-3)

Lowest (2,2)

b)

Farthest (-3,-3)

Closest (2,2)

Step-by-step explanation:

To solve this problem we will be using Lagrange multipliers.

a)

Let us find out first the restriction, which is the projection of the intersection on the XY-plane.

From x+y+z=12 we get z=12-x-y and replace this in the equation of the paraboloid:

\bf 12-x-y=x^2+y^2\Rightarrow x^2+y^2+x+y=12

completing the squares:

\bf x^2+y^2+x+y=12\Rightarrow (x+1/2)^2-1/4+(y+1/2)^2-1/4=12\Rightarrow\\\\\Rightarrow (x+1/2)^2+(y+1/2)^2=12+1/2\Rightarrow (x+1/2)^2+(y+1/2)^2=25/2

and we want the maximum and minimum of the paraboloid when (x,y) varies on the circumference we just found. That is, we want the maximum and minimum of  

\bf f(x,y)=x^2+y^2

subject to the constraint

\bf g(x,y)=(x+1/2)^2+(y+1/2)^2-25/2=0

Now we have

\bf \nabla f=(\displaystyle\frac{\partial f}{\partial x},\displaystyle\frac{\partial f}{\partial y})=(2x,2y)\\\\\nabla g=(\displaystyle\frac{\partial g}{\partial x},\displaystyle\frac{\partial g}{\partial y})=(2x+1,2y+1)

Let \bf \lambda be the Lagrange multiplier.

The maximum and minimum must occur at points where

\bf \nabla f=\lambda\nabla g

that is,

\bf (2x,2y)=\lambda(2x+1,2y+1)\Rightarrow 2x=\lambda (2x+1)\;,2y=\lambda (2y+1)

we can assume (x,y)≠ (-1/2, -1/2) since that point is not in the restriction, so

\bf \lambda=\displaystyle\frac{2x}{(2x+1)} \;,\lambda=\displaystyle\frac{2y}{(2y+1)}\Rightarrow \displaystyle\frac{2x}{(2x+1)}=\displaystyle\frac{2y}{(2y+1)}\Rightarrow\\\\\Rightarrow 2x(2y+1)=2y(2x+1)\Rightarrow 4xy+2x=4xy+2y\Rightarrow\\\\\Rightarrow x=y

Replacing in the constraint

\bf (x+1/2)^2+(x+1/2)^2-25/2=0\Rightarrow (x+1/2)^2=25/4\Rightarrow\\\\\Rightarrow |x+1/2|=5/2

from this we get

<em>x=-1/2 + 5/2 = 2 or x = -1/2 - 5/2 = -3 </em>

<em> </em>

and the candidates for maximum and minimum are (2,2) and (-3,-3).

Replacing these values in f, we see that

f(-3,-3) = 9+9 = 18 is the maximum and

f(2,2) = 4+4 = 8 is the minimum

b)

Since the square of the distance from any given point (x,y) on the paraboloid to (0,0) is f(x,y) itself, the maximum and minimum of the distance are reached at the points we just found.

We have then,

(-3,-3) is the farthest from the origin

(2,2) is the closest to the origin.

3 0
3 years ago
17d+32a=176 what is d?
Vlad [161]
Minus 32a from both sides
17d=176-32a
divide both sides by 17
d=\frac{176-32a}{17}
4 0
3 years ago
URGENTE, POR FAVOR !!!!!!!!!!!
il63 [147K]

Answer:

ffefecrvfffffStep-by-step explanation:

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6 0
3 years ago
Find the product. (x2 + 3x - 2)(x + 3) A. x3 + 6x2 + 7x - 6 B. x2 + 4x + 1 C. x3 + 3x2 + 1 D. x3 + 6x2 + 11x + 6
marta [7]

Answer:

Step-by-step explanation:

x^3 + 3x^2 - 2x + 3x^2 + 9x - 6

x^3 + 6x^2 + 7x - 6

the solution is A

8 0
3 years ago
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