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3 years ago

Amia went to a flower farm and picked f flowers. When

Mathematics

1 answer:

Salsk061 [2.6K]3 years ago

4 0

Answer:

4×22=x

She picked 88 flowers

Step-by-step explanation:

4×22=88

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Step-by-step explanation:

If 70 is 50% then 140 is 100%.

c is 75% of 140 = 105

a is 25% of 140 = 35

b is 100% of 140 =140

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2 years ago

Kahlia needs $468.75 to buy herself a new chromebook. She plans to wash 125 cars in her neigborhood to earn enough money to but

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468.75 / 125 = $3.75 per car wash.

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2 years ago

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What is the side length of square BDFH ?

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3 years ago

Describe the transformation done to y = 1/x to get y = 1/4x -12?

BabaBlast [244]

The transformation is done to y=1/x to get y=1/4x-12. We defined a turning point in the equation which will indicate the point in the graph.

We changed the form from y=x to y=x+c. In this case, we use 1/2 for the and constant C value which is -12.

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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:

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.

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3 years ago