Ask question

Ask question

All categories

4 years ago

12

5. Consider a cylindrical flowerpot with a radius 6 cm and a height of 11.5 cm. Calculate the lateral surface area of the painte

d pot.

1 answer:

Nina [5.8K]4 years ago

8 0

Answer: 660 cm²

Step-by-step explanation:

Surface area of a cylinder = 2πr² ± 2πrh = 2πr ( r + h )

π = 22/7

r = 6 cm

h = 11.5 cm

Lateral surface area = 2 x 22/7 x 6 ( 6 + 11.5 )

= 660 cm²

You might be interested in

Supply the missing information for each of the right triangles described below. In each example, θ = angle, O = opposite side, A

blagie [28]

Answer:

$\therefore O = 1.9096$

Step-by-step explanation:

Six trigonometric ratio:

$sin\theta=\frac{opposite}{hypoteuse}$

$cos\theta=\frac{adjacent}{hypotenuse}$

$tan \theta=\frac{opposite}{adjacent}$
$cosec\theta=\frac{hypotenuse}{opposite}$
$sec\theta=\frac{hypotenuse}{adjacent}$
$tan \theta=\frac{adjacent}{opposite}$

Given that,

θ = angle, O= opposite side, A= adjacent side and H = hypotenuse.

H=11, sin θ=0.1736

We know that,

$sin\theta=\frac{opposite}{hypoteuse}$

$\Rightarrow 0.1736=\frac{O}{11}$

$\Rightarrow O = 11 \times 0.1736$

$\Rightarrow O = 1.9096$

$\therefore O = 1.9096$

3 0

4 years ago

Please I’ll do anything I’m begging

VashaNatasha [74]

Answer:

14√2

Step-by-step explanation:

This is a 45-45-90 triangle

Because both of the angles are the same all sides, but the hypotonouse are the same

this means that the other length is 14 too

we can use a²+b²=c² to solve

14²+14²=c²

=

392=c²

√392=c

√392=14√2

4 0

4 years ago

Surface area of cylinder

steposvetlana [31]

Answer:

The answer is C. $256\pi$ $in^2$ .

Step-by-step explanation:

To solve for the surface area of this cylinder, start by using the cylinder surface area formula, which is $SA=2\pi rh+2\pi r^{2}$ . Next, plug in the information given from the problem into the formula, which will look like $SA=(2)(\pi )(8in)(8in)+(2)(\pi )(8in^{2})$ . Then, simplify the equation, which will look like $SA=(128in)(\pi) +(128in)(\pi)$ . The next step is to simplify the equation further, which will look like $SA= 256\pi$ $in^{2}$ . The final answer is $256 \pi$ $in^2$ .

8 0

3 years ago

alex won first place in the shot-put with a heave of 35 feet 4 inches. Griffin won second place with a heave of 32 feet 9 inches

Alex Ar [27]

Alex defeated Griffin by 2 feet and 5 inches. 2.5
How to solve: 35.4 - 32.9 = 2.5 :))

6 0

3 years ago

Use the upper and lower sums to approximate the area of the region using the given number of subintervals (of equal width)

Mashutka [201]

From the figure shown, the interval is divided into 5 equal parts making each subinterval to be 0.2.

Part A:

$y= \sqrt{1-x^2}$

The approximate the area of the region shown in the figure using the lower sums is given by:

$Area= [y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2]+[y(0.8)\times0.2] \\ +[y(1)\times0.2] \\ \\ =[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2]+[\sqrt{1-(0.6)^2}\times0.2] \\ +[\sqrt{1-(0.8)^2}\times0.2]+[\sqrt{1-(1)^2}\times0.2] \\ \\ =(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2)+(0\times0.2) \\ \\ =0.196+0.183+0.16+0.12=0.659$

Part B:

The approximate the area of the region shown in the figure using the lower sums is given by:

$Area= [y(0)\times0.2]+[y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2] \\ +[y(0.8)\times0.2] \\ \\ =[\sqrt{1-(0)^2}\times0.2]+[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2] \\ +[\sqrt{1-(0.6)^2}\times0.2] +[\sqrt{1-(0.8)^2}\times0.2] \\ \\ =(1\times0.2)+(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2) \\ \\ =0.2+0.196+0.183+0.16+0.12=0.859$

Part C:

The approximate area of the given region is given by

$Area= \frac{0.659+0.859}{2} = \frac{1.518}{2} =0.759$

7 0

4 years ago