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3 years ago

1. Find the circumference and area of circle with radius 5 cm.

Mathematics

2 answers:

Natali5045456 [20]3 years ago

6 0

Answer:

Hey There!

Let's solve...

First let's find out circumference

We know the formula is

$2 \pi r \\ \\ = 2 \times 5 \times 3.14cm \\ \\ = 10 \times 3.14cm \\ \\ = 31.4cm$

Now let's find out area...

The formula is

$\pi r^{2} \\ = 3.14 \times {5}^{2} \\ = 3.14 \times 25 \\ = 78.50 {cm}^{2}$

<h2>I hope it is helpful to you...</h2><h3>Cheers!_____________</h3>

SOVA2 [1]3 years ago

4 0

Answer:

3l.42cm

Step-by-step explanation:

C =2(3.14) × 5cm

C= 31.42cm

Area = pie r2

area =3.14 × 5cm ×5cm

area= 78.5cm

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Mr. Hayden's Bible class collected $11

ki77a [65]

This is the only way I can think of this mathematically. This took a good 20 minutes for me. You need to guess and check. Like try 17 quarters, 50 nickels, and 37 dimes etc. until you get 11.00. My numbers are mathematically correct but we both know you cant have 18.909 quarters so. lol

Answer:

Quarters: 18.909

Nickels: 49.818

Dimes: 37.818

Somewhere around those numbers.

Step-by-step explanation:

$11.00 total in coins

Nickel = $0.05

Dime = $0.10

Quarter = $0.25

(n*0.05) + (d*0.10) + (q*0.25)

Solve the equation for q.

$q=\frac{1}{2} d$

(n*0.05) + (d*0.10) +( $\frac{1}{2} d$ *0.25)

Simplify the expression.

9d + 2n = 440

d - n = -12

Multiply both sides of the equation by -9.

-9d + 9n = 108

9d + 2n = 440

Sum the equations vertically to eliminate at least one variable.

11n = 548

$n=\frac{548}{11}$

Substitute the given value of n into the equation d - n = -12

$d-\frac{548}{11} =-12$

$d=\frac{416}{11}$

Substitute the given value of n into the equation $q=\frac{1}{2} d$

$q=\frac{1}{2} *\frac{416}{11}$

$q=\frac{208}{11}$

Finally

$d=\frac{416}{11}$ d = 37.818

$n=\frac{548}{11}$ n = 49.818

$q=\frac{208}{11}$ q = 18.909

6 0

3 years ago

2. What is the vertex of the following parabola? *

sweet [91]

Answer:

Step-by-step explanation:

The answer to life the universe and everything is 42

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3 years ago

A ball is dropped from a height of 14 ft. The elasticity of the ball is such that it always bounces up one-third the distance it

Korvikt [17]

Answer:

Hello,

742/27 (ft)

Step-by-step explanation:

$h_1=14\\\\h_2=\dfrac{14}{3} \\\\h_3=\dfrac{14}{9} \\\\h_4=\dfrac{14}{27} \\\\$

$d=14+2*\dfrac{14}{3} +2*\dfrac{14}{9} +2*\dfrac{14}{27} \\=14*(1+\dfrac{1}{3}+\dfrac{2}{9} +\dfrac{2}{27} )\\=14*\dfrac{53}{27} \\=\dfrac{742}{27} \\$