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3 years ago

The graph shows the cost per day for renting a car. Which statement is true?

Mathematics

2 answers:

WINSTONCH [101]3 years ago

7 0

C. is your answer. If the graph had a negative rate of change, the numbers would be smaller and might even go into the negatives

Natalka [10]3 years ago

6 0

Answer:

Step-by-step explanation:

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A canoe rental service charges a $20 transportation fee and $30 an hour to rent a canoe. Find the slope and y-intercept for the

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2 years ago

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2 years ago

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-5(w+4) +8< - 42 need a quick answer

Nesterboy [21]

Answer:

$w > 6$

Step-by-step explanation:

Let's solve your inequality step-by-step.

−5(w+4)+8<−42

Step 1: Simplify both sides of the inequality.

−5w−12<−42

Step 2: Add 12 to both sides.

−5w−12+12<−42+12

−5w<−30

Step 3: Divide both sides by -5.

$\frac{-5w}{-5} < \frac{-30}{-5}$

$w > 6$

<u>Answer: </u>

<u /> $w > 6$

Thanks!

- Eddie

7 0

1 year ago

This is urgent, please answer!

Alex17521 [72]

1/2 = 0.5

12 & 1/2 = 12 + 1/2 = 12 + 0.5 = 12.5

4 & 1/2 = 4 + 1/2 = 4 + 0.5 = 4.5

Refer to the diagram below. Note how I divided the figure into two rectangles.

The larger red rectangle is 12.5 feet by 10 feet. It has area 12.5*10 = 125 square feet.

The smaller blue rectangle has dimensions 4.5 feet by 5 feet (the 5 is from 15-10 = 5), so it has area 4.5*5 = 22.5 square feet

Now add up those individual areas to get the total area

125+22.5 = 147.5

Then convert that to a mixed number

147.5 = 147 + 0.5 = 147 + 1/2 = 147 & 1/2

<h3>The correct area is 147 & 1/2 feet</h3>

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:

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