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-Dominant- [34]
2 years ago
7

What is the Error 1.38 ÷ 3

Mathematics
1 answer:
cestrela7 [59]2 years ago
6 0
You are................
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Select "Yes" or "No" to indicate whether the ordered pair is on the graph of the function f(x)=−16x+1 .
alexandr402 [8]
To check if the pair is on the graph, plug the x value into the x in the equation, and solve. If the solution to the equation matches the y value from your ordered pair, it’s on the graph, so mark “yes”. If the solution isn’t the same, it’s not on the graph, so mark “no”.
8 0
3 years ago
Please help I need it
ivolga24 [154]

Answer:

<h2>Yes. It's a right triangle.</h2>

Step-by-step explanation:

If a ≤ b < c are the length of the sides of a right triangle, then

a² + b² = c².

We have:

a =30 ft, b = 40 ft and c = 50 ft.

Check the equality:

L = 30² + 40² = 900 + 1600 = 2500

R = 50² = 2500

L = R

5 0
2 years ago
Find the maximum volume of a rectangular box that is inscribed in a sphere of radius r.
zvonat [6]

Answer:

The maximum volume of a box inscribed in a sphere of radius r is a cube with volume \frac{8r^3}{3\sqrt{3}}.

Step-by-step explanation:

This is an optimization problem; that means that given the constraints on the problem, the answer must be found without assuming any shape of the box. That feat is made through the power of derivatives, in which all possible shapes are analyzed in its equation and the biggest -or smallest, given the case- answer is obtained. Now, 'common sense' tells us that the shape that can contain more volume is a symmetrical one, that is, a cube. In this case common sense is correct, and the assumption can save lots of calculations, however, mathematics has also shown us that sometimes 'common sense' fails us and the answer can be quite unintuitive. Therefore, it is best not to assume any shape, and that's how it will be solved here.

The first step of solving a mathematics problem (after understanding the problem, of course) is to write down the known information and variables, and make a picture if possible.

The equation of a sphere of radius r is x^2 + y^2 + z^2=r^2. Where x, y and z are the distances from the center of the sphere to any of its points in the border. Notice that this is the three-dimensional version of Pythagoras' theorem, and it means that a sphere is the collection of coordinates in which the equation holds for a given radius, and that you can treat this spherical problem in cartesian coordinates.

A box that touches its corners with the sphere with arbitrary side lenghts is drawn, and the distances from the center of the sphere -which is also the center of the box- to each cartesian axis are named x, y and z; then, the complete sides of the box are measured  2x,  2y and 2z. The volume V of any rectangular box is given by the product of its sides, that is, V=2x\cdot 2y\cdot 2z=8xyz.

Those are the two equations that bound the problem. The idea is to optimize V in terms of r, therefore the radius of the sphere must be introduced into the equation of the volumen of the box so that both variables are correlated. From the equation of the sphere one of the variables is isolated: z^2=r^2-x^2 - y^2\quad \Rightarrow z= \sqrt{r^2-x^2 - y^2}, so it can be replaced into the other: V=8xy\sqrt{r^2-x^2 - y^2}.

But there are still two coordinate variables that are not fixed and cannot be replaced or assumed. This is the point in which optimization kicks in through derivatives. In this case, we have a cube in which every cartesian coordinate is independent from each other, so a partial derivative is applied to each coordinate independently, and then the answer from both coordiantes is merged into a single equation and it will hopefully solve the problem.

The x coordinate is treated first: \frac{\partial V}{\partial x} =\frac{\partial 8xy\sqrt{r^2-x^2 - y^2}}{\partial x}, in a partial derivative the other variable(s) is(are) treated as constant(s), therefore the product rule is applied: \frac{\partial V}{\partial x} = 8y\sqrt{r^2-x^2 - y^2}  + 8xy \frac{(r^2-x^2 - y^2)^{-1/2}}{2} (-2x) (careful with the chain rule) and now the expression is reorganized so that a common denominator is found \frac{\partial V)}{\partial x} = \frac{8y(r^2-x^2 - y^2)}{\sqrt{r^2-x^2 - y^2}}  - \frac{8x^2y }{\sqrt{r^2-x^2 - y^2}} = \frac{8y(r^2-2x^2 - y^2)}{\sqrt{r^2-x^2 - y^2}}.

Since it cannot be simplified any further it is left like that and it is proceed to optimize the other variable, the coordinate y. The process is symmetrical due to the equivalence of both terms in the volume equation. Thus, \frac{\partial V}{\partial y} = \frac{8x(r^2-x^2 - 2y^2)}{\sqrt{r^2-x^2 - y^2}}.

The final step is to set both partial derivatives equal to zero, and that represents the value for x and y which sets the volume V to its maximum possible value.

\frac{\partial V}{\partial x} = \frac{8y(r^2-2x^2 - y^2)}{\sqrt{r^2-x^2 - y^2}} =0 \quad\Rightarrow r^2-2x^2 - y^2=0 so that the non-trivial answer is selected, then r^2=2x^2+ y^2. Similarly, from the other variable it is obtained that r^2=x^2+2 y^2. The last equation is multiplied by two and then it is substracted from the first, r^2=3 y^2\therefore y=\frac{r}{\sqrt{3}}. Similarly, x=\frac{r}{\sqrt{3}}.

Steps must be retraced to the volume equation V=8xy\sqrt{r^2-x^2 - y^2}=8\frac{r}{\sqrt{3}}\frac{r}{\sqrt{3}}\sqrt{r^2-\left(\frac{r}{\sqrt{3}}\right)^2 - \left(\frac{r}{\sqrt{3}}\right)^2}=8\frac{r^2}{3}\sqrt{r^2-\frac{r^2}{3} - \frac{r^2}{3}} =8\frac{r^2}{3}\sqrt{\frac{r^2}{3}}=8\frac{r^3}{3\sqrt{3}}.

6 0
3 years ago
Line KL has an equation of a line y = 4x + 5. Which of the following could be an equation for a line that is perpendicular to li
djyliett [7]
To find a perpendicular slope (or line), the slope (in this case 4x) must be the opposite sign and its reciprocal, which is basically the fraction flipped upside down. Since 4 is technically 4/1, that fraction flipped is 1/4. And since you need to flip the sign too, instead of it being a positive number, it's negative. Your answer is -1/4x-8
3 0
3 years ago
Read 2 more answers
If an architect wants to draw a plan for a house that is 75 ft by 80 ft, and his paper is 10 in. by 17 in., what is the best sca
drek231 [11]
The best scale factor to make use of every inch of the paper would be that 1 inch represents 7.5 feet.
7 0
2 years ago
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