Help me please again

the perimeter of the circular base of a cone is 132 cm and its vertical height is 28 cm find the curved surface area total surfa

Elden [556K]

Answer:

i. Curved surface area = 2310 $cm^{2}$

ii. Total surface area = 3696 $cm^{2}$

iii. volume = 12936 $cm^{3}$

Step-by-step explanation:

Perimeter of the circular base = circumference of the circle = 132 cm

circumference of a circle = 2 $\pi$ r

132 = 2 $\pi$ r

r = $\frac{132}{2\pi }$

= $\frac{66}{\pi }$

r = 66 x $\frac{7}{22}$

= 3 x 7

= 21

radius = 21 cm

vertical height, h = 28 cm

Thus applying the Pythagoras theorem,

$l^{2}$ = $h^{2}$ + $r^{2}$

= $28^{2}$ + $21^{2}$

= 784 + 441

= 1225

l = $\sqrt{1225}$

l = 35 cm

The slant height is 35 cm.

i. Curved surface area = $\pi$ rl

= $\frac{22}{7}$ x 21 x 35

= 22 x 3 x 35

= 2310

curved surface area of the cone is 2310 $cm^{2}$ .

ii. Total surface area = $\pi r^{2}$ + $\pi$ rl

= $\pi$ r(r + l)

= $\frac{22}{7}$ x 21 (21 + 35)

= 22 x 3 x 56

= 3696

The total surface area of the cone is 3696 $cm^{2}$ .

iii. volume of a cone = $\frac{1}{3}$ $\pi r^{2}$ h

= $\frac{1}{3}$ x $\frac{22}{7}$ x $21^{2}$ x 28

= 12936

The volume of the cone is 12936 $cm^{3}$

5 0

3 years ago

He mean potassium content of a popular sports drink is listed as 140 mg in a 32-oz bottle. Analysis of 20 bottles indicates a sa

mezya [45]

Answer: (A) H0: μ = 140 mg vs. H1: μ ≠ 140 mg

Step-by-step explanation:

Usually null hypothesis represents the claim that the values associated to the groups being tested have no statistical difference but alternative hypothesis supports the claim that there is statistical difference.

Let $\mu$ be the population mean .

We are given that the mean potassium content of a popular sports drink is listed as 140 mg in a 32-oz bottle.

i.e. Null hypothesis : $H_0:\mu=140$

Alternative hypothesis for two tail hypothesis has sign (≠).

i.e. Alternative hypothesis : $H_1 : \mu\neq140$

∴ The hypotheses for a two-tailed test of the claimed potassium content:

H0: μ = 140 mg vs. H1: μ ≠ 140 mg

3 0

3 years ago

1000 cylindrical metal bars of length 120 m and diameter 0.2 m , as shown below, are to be used in constructing a bridge. He bar

Troyanec [42]

Answer:

the number of of litres of paint to the nearest litre for 1000 bars is 1 litre

Step-by-step explanation:

Given that:

1000 cylindrical metal bars have :

a length of 120 m

and a diameter of 0.2 m

The volume of the cylindrical bar can be calculated by the formula:

$V= \pi r^2 h$

where ;

radius $r=\dfrac{d}{2}$

$r = \dfrac{0.2}{2}$

r = 0.1 m

the length of the cylindrical metal bars is equal to the height of the metal bar = 120 m

Thus the volume can now be calculated as follows:

$V= \pi* 0.1^2*120$

V = 3.77 m³

So; from the given information.

One litre of paint will cover 4m³

and we want to find how many litres of paint will be used for the 1000 cylindrical bars.

If 1 litre = 4m³

Then x litre = 3.77 m³

4x = 1 × 3.77

$x = \dfrac{1*3.77 }{4}$

x = 0.9425 litres

x = 1 litres to the nearest litres.

Thus ; the number of of litres of paint to the nearest litre for 1000 bars is 1 litre

4 0

4 years ago

The Springfield School band has 300 sweatshirts to sell at the fund-raiser.A survey shows that 2 out of 7 students in the school

nydimaria [60]

260 students is the prediction for how many sweatshirts will be sold so no they will not sell all of them

5 0

4 years ago

Which statement regarding the diagram is true?

Svetlanka [38]

You need to post the diagram!

8 0

2 years ago