All categories

3 years ago

2) Find the area and perimeter of a semi -circular disc whose radius is 10cm. (take π = 3.14)

Mathematics

2 answers:

SVETLANKA909090 [29]3 years ago

5 0

Answer:

=3.14×10x10

=3.14×100cm2

=3.14cm2

<h3>Perimeter = 2 pi r</h3>

=2 ×3.14×10

=6.28cm

Or =Pi d

= 3.14 ×20cm

=62.8cm

R means radius

D means diameter

radius is half of a diameter e.g radius is10 and diameter is 20

I can recommend you watching yt videos it helped me alot

hoo this helps:)

a_sh-v [17]3 years ago

4 0

Answer:

i d k

Step-by-step explanation:

You might be interested in

What are ALL the solutions 16x^2-12x=54

nikdorinn [45]

Answer:x=9/4 and x=-3/2

Step-by-step explanation:

4 0

3 years ago

PLEASE HELP RIGHT AWAY, <br> THANK U SO MUCH!! <3

vredina [299]

Answer:

1. Ans;

${a}^{7} \times {a}^{12} = {a}^{7 + 12} = {a}^{19}$

__o_o__

2. Ans;

$\frac{ {h}^{6} }{ {h}^{2} } = {h}^{4}$

__o_o__

3. Ans;

${e}^{6} \times {e}^{6} = {e}^{6 + 6} = {e}^{12}$

__o_o__

4.Ans;

$\frac{ {w}^{9} }{ {w} } = {w}^{8}$

__o_o__

5. Ans;

${g}^{2} \times {h}^{3} \times {g}^{2} \times h = {g}^{2 + 2} \times {h}^{3 + 1} = {g}^{4} {h}^{4}$

__o_o__

6.Ans;

$\frac{10 {a}^{12} }{ 6{a}^{3} } = \frac{5 {a}^{9} }{3}$

__o_o__

7.Ans;

$({m}^{4} )^{5} = {m}^{4 \times 5} = {m}^{20}$

__o_o__

8.Ans;

${(3 {a}^{2} })^{3} = ( {3})^{3} ( {a}^{2} ) ^{3} = 27 {a}^{6}$

__o_o__

9.Ans;

${( \frac{c}{k} })^{4} = \frac{ {c}^{4} }{ {k}^{4} }$

__o_o__

10.Ans;

${( \frac{8x}{9} })^{2} = \frac{( {8x})^{2} }{ ({9})^{2} } = \frac{64 {x}^{2} }{81}$

__o_o__

11.Ans;

$4 {c}^{2} \times {c}^{3} = 4 {c}^{2 + 3} = 4 {c}^{5}$

__o_o__

12.Ans;

${( - 9 {h}^{4} {k}^{6} })^{2} = {( - 9)}^{2} {( {h}^{4} })^{2} {( {k}^{6} })^{2} = 81 \: \: {h}^{8} {k}^{12}$

__o_o__

13.Ans;

$( { \frac{2d}{3e} })^{3} = \frac{ {(2d)}^{3} }{ ({3e)}^{3} } = \frac{8 {d}^{3} }{27 {e}^{3} }$

__o_o__

14. Ans;

${( \frac{4}{ {w}^{5} }) }^{3} = \frac{ ({4)}^{3} }{( {w}^{5} ) ^{3} } = \frac{64}{ {w}^{15} }$

_____o___o_____

$\frac{5 {v}^{4} \times 4 {v}^{5} }{2 {v}^{3} } = \frac{20 {v}^{4 + 5} }{2 {v}^{3} } = \frac{20 {v}^{9} }{2 {v}^{3} } = 10 {v}^{6}$

_o_o_

$( \frac{3 {f}^{4}h}{4 {h}^{9} } ) ^{3} = \frac{( {3})^{3} ( {f}^{4})^{3} ({h})^{3} }{ {(4)}^{3} ( {h}^{9})^{3} } \\ \\ \\ = \frac{27 {f}^{4 \times 3} {h}^{3} }{64 {h}^{9 \times 3} } = \frac{27 {f}^{12} {h}^{3} }{64 {h}^{27} } \\ \\ \\ = \frac{27 {f}^{12} }{64 {h}^{24} }$

I hope I helped you^_^

5 0

3 years ago

Resuelve las ecuaciones a+(-8/3)=23, 2y+(-5/4)=0,75, 18*=1/3, m-5=2/7<br><br> AYUDA POR FAVOR

Marina86 [1]

Answer:

a bueno bro pa saber.....

4 0

3 years ago

Find an equation of the parabola with focus (2, −3) and directrix y = 1.

Likurg_2 [28]

Answer:

Step-by-step explanation:

The correct option is option (C).

The required equation of the parabola is

Step-by-step explanation:

Formula:

The distance of a point (x₁,y₁) from a line ax+by+c=0 is

The distance between two points (x₁,y₁) and (x₂,y₂) is

Given that, the focus of the parabola is (2,3) and directrix y= - 1.

We know that, a parabola is the locus of points that equidistant from the directrix and the focus.

Let any point on the parabola be P(x,y) .

The distance of P from the directrix is

=y+1

The distance of the point P from the focus (2,3) is

According to the problem,

The required equation of the parabola is

5 0

3 years ago

Gravel is being dumped from a conveyor belt at a rate of 40 ft3/min, and its coarseness is such that it forms a pile in the shap

ikadub [295]

The height is increasing at the rate of (6/5π) ft/min.

It is given that Gravel is being dumped from a conveyor belt at a rate of 40 ft3/min. Also, its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal.

For this, we need to find the volume of a cone. Volume is a scalar quantity expressing the amount of three-dimensional space enclosed by a closed surface.

The volume of a cone is - V=1/3hπr²

Learn more about integers here: brainly.com/question/20521181

#SPJ4

4 0

2 years ago