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2 years ago

Find the surface area. round to the nearest hundredth when necessary.

Mathematics

1 answer:

kherson [118]2 years ago

8 0

Answer:

We have the surface area of the cylinder as 3,016.32 in^2

Step-by-step explanation:

What we have here is a cylinder of height 14 inches and diameter 32 inches

Mathematically we can calculate the surface area of a cylinder using the formula;

A = 2 * pi * r(r + h)

where r is diameter/2 = 32/2 = 16 inches

h = 14 inches

pi = 3.142

thus, we have it that;

A = 2 * 3.142 * 16(16 + 14)

A = 3016.32 in^2

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How do i solve -3=-2x+1 by graphing?

KonstantinChe [14]

Answer:

x = 2

Step-by-step explanation:

-3 = -2x + 1

minus 1 to the both sides

-3 -1 = -2x

-4 = -2x

divide both sides by -2

x = 2

8 0

3 years ago

A special liquid is held in a tank described as (x 2 + y 2 ) ≤ z ≤ 1 in a Cartesian coordinate system. Assume that the density o

vivado [14]

Since $\rho=\dfrac mV$ (density = mass/volume), we can get the mass/weight of the liquid by integrating the density $\rho(x,y,z)$ over the interior of the tank. This is done with the integral

$\displaystyle\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{x^2+y^2}^1(2-z^2)\,\mathrm dz\,\mathrm dy\,\mathrm dx$

which is more readily computed in cylindrical coordinates as

$\displaystyle\int_0^{2\pi}\int_0^1\int_{r^2}^1(2-z^2)r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\boxed{\frac{3\pi}4}$

7 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

3 years ago

What is the volume of this shape?

Stells [14]

Answer:

Step-by-step explanation:

3 x 3 x 4 =

9 x 4 =

6 0

2 years ago

Read 2 more answers

!!!!PLEASE HELP!!!

Law Incorporation [45]

Answer: 346.4

Step-by-step explanation:

The whole triangle ABD is a 30-60-90 right triangle. In 30-60-90 triangles, the smallest side is x, the middle side is x $\sqrt{3}$ , and the largest side/hypotenuse is 2x.

Side AB is the smallest side and therefore x=300. Side BD is the middle side and so we can use x $\sqrt{3}$ and we get BD=300 $\sqrt{3}$ =519.615

The question is asking for CD, and we know that CD can be equal to BD minus BC since that would isolate CD

Triangle ABC is also a 30-60-90 triangle but the 300 feet is the middle side in this triangle. So we can do 300=x $\sqrt{3}$ and divide 300 by radical 3 and get that x=173.205

Side BC is the smallest side of this triangle so it would be x and x is 173.205.

Now that we have the whole side BD and the part of it BC. We can subtract BD by BC (519.615-173.205) and we get that CD=346.4

5 0

2 years ago