All categories

3 years ago

The price to ship a box varies directly as the weight of the box in pounds. A box that weighs 10 pounds costs $10.90 to ship.

1 answer:

3 0

Well we can say 10.90/10 = the cost per pound to ship the item
we can infer this from the above text so
10.90/10 = 1.09.
So 1.09 * 37 = 40.33
$40.33
or 40 dollars and 33 cents to ship the second box with a weight of 37lb (pounds).

You might be interested in

I am building a fence in my backyard. I have 120 nails, and I need to put 4 nails

Vsevolod [243]

120/4 is 30
You'll be able to put 4 nails thirty times.
You put them every 6 feet
So, you have 5 feet before running out.

8 0

3 years ago

Read 2 more answers

What is the slope of the line <br><br>

prisoha [69]

Answer :

The slope of the line is 0

3 0

3 years ago

Use a table to compute the product below.

Daniel [21]

(2x+2)(10x^2 - 6x + 8) =
2x(10x^2 - 6x + 8) + 2(10x^2 - 6x + 8) =
20x^3 - 12x^2 + 16x + 20x^2 - 12x + 16 =
20x^3 + 8x^2 + 4x + 16 <==

5 0

3 years ago

The path of a projectile launched from a 20-ft-tall tower is modeled by the equation y = -5x2 + 40x + 20. What is the maximum he

Otrada [13]

Answer:

30.49 m

Step-by-step explanation:

To obtain the maximum height, we solve for the value x when dy/dx = 0.

Since, y = -5x² + 40x + 20

dy/dx = d[-5x² + 40x + 20]/dx

dy/dx = -10x + 40

Since dy/dx = 0,

-10x + 40 = 0

-10x = -40

x = -40/-10

x = 4

Substituting x = 4 into the equation for y, we have

y = -5x² + 40x + 20

y = -5(4)² + 40(4) + 20

y = -5(16) + 160 + 20

y = -80 + 160 + 20

y = 80 + 20

y = 100 ft

Since y is in feet, we convert to meters.

Since 1 m = 3.28 ft, 100 ft = 100 ft × 1 m/3.28 ft = 30.49 m

So, the maximum height, in meters reached by the projectile is 30.49 m

3 0

2 years ago

The foci of an ellipse are located at (0,6) and (0,-6). The sum of the distances from any point on the ellipse to the foci is 20

zlopas [31]

Answer:

Step-by-step explanation:

We have to sum up some of the major points of an ellipse before we can get to the answer here. First of all, the foci are ALWAYS on the major axis. Because we are given the coordinates of the foci as (0, 6) and (0, -6) we know the foci are on the y axis (because x = 0 in our foci coordinates) and the ellipse will have the equation:

$\frac{x^2}{b^2} +\frac{y^2}{a^2} =1$

This is the equation because: 1. the x and y DO NOT MOVE IN THE EQUATION FOR AN ELLIPSE! THE X WILL ALWAYS BE IN THE FIRST FRACTION! and 2. A is ALWAYS larger than b in an ellipse. Therefore, if the ellipse is elongated on the y-axis, the a goes under the y axis because the y axis is longer than the x-axis when the foci lie on the the y-axis. Did I mention that the foci ALWAYS lie on the major axis (the longer one)!? They do! And the distance from the center to the foci is a length of "c". Ours is 6. Summing up, ours is a vertically stretched ellipse because the foci lie on the y-axis (the vertical axis) and the foci always lie on the major axis. Hence, the y-axis is the major axis. And the center is directly in between the 2 foci. So since our foci are at (0, 6) and (0, -6), our center is at (0, 0).

The vertex also lie on the major axis, at "a". They are a bit further from the center than the foci are. The co-vertex is on the minor, or shorter, axis, at "b". There is a very important thing to know about the foci of an ellipse that will help us solve your problem, which is actually quite a problem indeed!

Pick a point on an ellipse. The distance from this point to one focal point, let's call that distance d1, when added to the distance from this same point to the other focal point, let's call that distance d2, will ALWAYS BE THE SAME. Our distances, d1 + d2, we are told equal 20.

Let's see if we can put some of this together now and solve your problem.

Because a is the distance from the center to one vertex and a is on the major axis, then the length of the major axis is equal to 2a.

Likewise, because b is the distance from the center to one co-vertex and b is on the minor axis, then the length of the minor axis is equal to 2b.

If d1 + d2 = 20 and d1 + d2 always has the same length as the major axis, then:

2a = 20 and

a = 10.

We already know that c = 6.

To find b, then, we need the equation to find the foci of an ellipse which is

$c^2=a^2-b^2$