Ask question

Ask question

All categories

3 years ago

15

Inga knows that the radius of a certain cone is 4 cm and its volume is about 134 cubic centimeters. Which is the best estimate o

f the cone's height?

1 answer:

Veronika [31]3 years ago

8 0

Answer:

h=8cm

Step-by-step explanation:

V= πr^2h/3

Where

V=volume of a cone

π=pi= 3.14

r= radius=4cm

h= height = ?

V= πr^2h/3

134= 3.14*4*4*h/3

134= 50.24h/3

Divide both sides by 50.24

h/3= 134 / 50.24

h/3= 2.67

Cross product

h= 2.67*3

h= 8.01

Approximately h= 8cm

The best estimate of the cone's height is 8cm

You might be interested in

URGENT: WILL MARK BRAINLIEST

saveliy_v [14]

Answer:

please wait for 10 minutes I will give your answer

5 0

4 years ago

Read 2 more answers

Which is equal to -214°?

Neporo4naja [7]

Answer: -107pi/45 radians

5 0

3 years ago

Solve the following equations;-<br><br> x - 9 /√x + 3 =1

fgiga [73]

Answer:

13 , 6.

Step-by-step explanation:

A equation is given to us and we need to find out the value of x . The given equation to us is ,

$\sf\implies \dfrac{ x -9}{\sqrt{x+3}}= 1$

Cross multiply ,

$\sf\implies x - 9 =\sqrt{ x +3}$

Squaring both sides ,

$\sf\implies (x - 9)^2 = x + 3$

Simplify the whole square ,

$\sf\implies x^2+9^2-2.9.x = x +3 \\$

$\sf\implies x^2+81 - 18x = x -3$

Add 3 and subtract x on both sides ,

$\sf\implies x^2 -18x-x +81-3=0$

Simplify ,

$\sf\implies x^2 -19x +78=0$

Split the middle term of the quadratic equation,

$\sf\implies x^2-13x-6x+78=0$

Take out common ,

$\sf\implies x( x -13)-6(x-13)=0$

Take out ( x -13 ) as common ,

$\sf\implies ( x -13)(x-6)=0$

Equate both factors to 0 ,

$\sf\implies \boxed{\pink{\sf x = 13,6}}$

<u>Hence</u><u> the</u><u> </u><u>value</u><u> of</u><u> </u><u>x</u><u> is</u><u> </u><u>1</u><u>3</u><u> </u><u>or </u><u>6</u><u> </u><u>.</u>

7 0

3 years ago

Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an

Leya [2.2K]

Parameterize the lateral face $T_1$ of the cylinder by

$\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v$

where $0\le u\le2\pi$ and $0\le v\le3$ , and parameterize the disks $T_2,T_3$ as

$\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)$
$\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)$

where $0\le r\le2$ and $0\le\theta\le2\pi$ .

The integral along the surface of the cylinder (with outward/positive orientation) is then

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$

7 0

3 years ago

-5х + Зу = -8<br> 3х - 7y = -3<br> Ordered pair

tatiyna

Answer:

1. y= 2x-5

2. y= x+3

3. y=2x+7

Step-by-step explanation:

6 0

3 years ago