All categories

3 years ago

Pls helpppppppppppppppp

Mathematics

2 answers:

lorasvet [3.4K]3 years ago

6 0

The volume, fr I don’t rly know sry

Viefleur [7K]3 years ago

5 0

The formula for volume is (length x width c height) since it is a cube all the sides should equal the same length.

You might be interested in

The table shows the amount of sugar needed to make lemonade. Are the quantities proportional? Explain which method you used and

Klio2033 [76]

Answer: Yes

Step-by-step explanation:

For the first sugar and servings part, 4 for sugar, 2 for servings.

The ratio for this example above it for sugars to servings it 2:1. So every 2 sugars is every 1 serving and this applies to every other of the sugars and servings in the table.

7 0

2 years ago

Read 2 more answers

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:

Highest (-3,-3)

Lowest (2,2)

Farthest (-3,-3)

Closest (2,2)

Step-by-step explanation:

To solve this problem we will be using Lagrange multipliers.

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

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

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

Right before a snow storm the hardware store sold

eduard

Answer:

Step-by-step explanation:

60/15=4

7 0

3 years ago

Which equation represents a line with a slope of zero?

elena55 [62]

Y=4 because it only has a y-intercept

3 0

2 years ago

1) <1 and <2 are supplementary angles. If m<1 = (4x)º and m<2 = (8x), then

Andru [333]

Answer:

i think the answer =x€R\(0)

8 0

2 years ago

Read 2 more answers