PLZ I NEED HELP WITH THIS QUESTION

xxTIMURxx [149]

Answer:

Step-by-step explanation:

If we assume that the angle at A is 90 degrees.. which it does seem like it would be. then use that the sum of the angles around the inside of a right triagle is 180 . so you have

180 = 37 + 90 + x

53° = x

does that help?

4 0

3 years ago

The surface area of a right circular cone of radius r and height h is S = πr√ r 2 + h 2 , and its volume is V = 1 3 πr2h. What i

kirill115 [55]

Answer:

Required largest volume is 0.407114 unit.

Step-by-step explanation:

Given surface area of a right circular cone of radious r and height h is,

$S=\pi r\sqrt{r^2+h^2}$

and volume,

$V=\frac{1}{3}\pi r^2 h$

To find the largest volume if the surface area is S=8 (say), then applying Lagranges multipliers,

$f(r,h)=\frac{1}{3}\pi r^2 h$

subject to,

$g(r,h)=\pi r\sqrt{r^2+h^2}=8\hfill (1)$

We know for maximum volume $r\neq 0$ . So let $\lambda$ be the Lagranges multipliers be such that,

$f_r=\lambda g_r$

$\implies \frac{2}{3}\pi r h=\lambda (\pi \sqrt{r^2+h^2}+\frac{\pi r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}r h= \lambda (\sqrt{r^2+h^2}+\frac{ r^2}{\sqrt{r^2+h^2}})\hfill (2)$

And,

$f_h=\lambda g_h$

$\implies \frac{1}{3}\pi r^2=\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}$

$\implies \lambda=\frac{r\sqrt{r^2+h^2}}{3h}\hfill (3)$

Substitute (3) in (2) we get,

$\frac{2}{3}rh=\frac{r\sqrt{R^2+h^2}}{3h}(\sqrt{R^2+h^2+}+\frac{r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}rh=\frac{r}{3h}(2r^2+h^2)$

$\implies h^2=2r^2$

Substitute this value in (1) we get,

$\pi r\sqrt{h^2+r^2}=8$

$\implies \pi r \sqrt{2r^2+r^2}=8$

$\implies r=\sqrt{\frac{8}{\pi\sqrt{3}}}\equiv 1.21252$

Then,

$h=\sqrt{2}(1.21252)\equiv 1.71476$

Hence largest volume,

$V=\frac{1}{3}\times \pi \times\frac{\pi}{8\sqrt{3}}\times 1.71476=0.407114$

3 0

3 years ago

The product of two consecutive integers is 420

Olenka [21]

(a) * (a+1) = 420
a^2 + a = 420
a^2 + a - 420 = 0
(a+21)(a-20) = 0
a = -21 OR 20

The two integers can be -21 and -20, OR 20 and 21.

6 0

3 years ago

A circle has a circumference of 18. It has an arc of 6. What is the central angle of the arc, in degrees?

irga5000 [103]

Answer:

Step-by-step explanation:

C=2πr

18=2×3.14×r

18=6.28r

r=2.88

Arc of a circle

A=®/360°πr

6=®/360°×3.14×r

®=239°

6

3 0

3 years ago

Read 2 more answers

[10 points] A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimens

MissTica

Answer:

6 inches square by 3 inches high

Step-by-step explanation:

For a given surface area, the volume of an open-top box is maximized when it has the shape of half a cube. If the area were than of the whole cube, it would be 216 in² = 6×36 in².

That is, the bottom is 6 inches square, and the sides are 3 inches high.

_____

Let x and h represent the base edge length and box height, respectively. Then we have ...

x² +4xh = 108 . . . . box surface area

Solving for height, we get ...

h = (108 -x²)/(4x) = 27/x -x/4

The volume is the product of base area and height, so is ...

V = x²h = x²(27/x -x/4) = 27x -x³/4

We want to maximize the volume, so we want to set its derivative to zero.

dV/dx = 0 = 27 -(3/4)x²

x² = (4/3)(27) = 36

x = 6

h = 108/x² = 3

The box is 6 inches square and 3 inches high.

_____

Comment on maximum volume, minimum area

In the general case of an open-top box, the volume is maximized when the cost of the bottom and the cost of each pair of opposite sides is the same. Here, the "cost" is simply the area, so the area of the bottom is 1/3 the total area, 36 in².

If the box has a closed top, then each pair of opposite sides will have the same cost for a maximum-volume box. If costs are uniform, the box is a cube.

7 0

3 years ago