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xenn [34]
4 years ago
7

Find the perimeter and area. (Plz help with this problem)

Mathematics
1 answer:
PIT_PIT [208]4 years ago
7 0

The first step to solving this problem is to find the area and circumference of the half circle using these equations:
d/\frac{1}{2} =r r*\pi=C and C/\frac{1}{2}= a
d= diameter
r= radius
C= circumference
a= answer

So the the "d" of the half circle is 10, and half of 10 is
r=5 . Because we are solving for only half of the circle this is what happens:
5*\pi =15.71
15.71*\frac{1}{2} =7.86
Now we just add the other sides together: 10+19+19+7.86=55.86m
The Perimeter is 55.86

Now for the area we must use a similar equation for the half circle: Area=\pi r^2
5^2=2525*\pi =78.54

This is the area of the whole circle so we have to divide 78.54 by half to get our answer, which is 39.27

10*19=190 This is the area of the rectangle
190+ 39.27=229.27 Is the area of the whole shape

Perimeter= 55.86
Area= 229.27

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PLEASE HELP VERY URGENT WILL GIVE BRAINILY PLEASE I BEG YOU PLEASE
yuradex [85]

Answer: 20.5 units

Step-by-step explanation:

P=Perimeter

JI=3-(-3)

JI=3+3

JI=6

IK=7-1

IK=6

JK=(6^2+6^2)^1/2   ==> Distance Formula

JK=(36+36)^1/2

JK=(2*36)^1/2

JK=6(2)^1/2

P=JI+IK+JK

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P=20.485 units

P=20.5 units

5 0
1 year ago
Use euler's theorem to find a number x between 0 and 28 with x85 congruent to 6 modulo 35.
Elan Coil [88]
Problem: find 0 ≤ x ≤ 28 such that x^85 ≡ 6 modulo 35.

By Fermat-Euler theorem:
If a and n are coprime, i.e. (a,n)=1, then
a^phi(n) ≡ 1 mod n  
where phi(n)=totient function, the number of positive integers less than n that is coprime with n.

for n=35, phi(35)=24 calculated as follows:
There are 10 positive integers from 1 to 34 which are NOT coprime with 35, namely {5,7,10,14,15,20,21,25,28,30}.Therefore phi(35)=34-10=24

From Fermat-Euler theorem, 
x^(phi(35) = x^24 ≡ 1 modulo 35  since (24,35)=1, i.e. 24 and 35 are coprime.
=>
x^12 ≡ ± 1 modulo 35. ...........(1)

and 

x^85 ≡ x^(85-3*24) ≡ x^(85-72) ≡ x^(13) ≡ 6 mod 35 ............(2)

Substituting (1) in (2)
x^(12)*x ≡ 6 mod 35
=>
(+1)*x = 6 mod 35   or  (-1)*x ≡ 6 mod 35
x ≡ 6 mod 35           x ≡ -6 mod 35   (rejected)

=>  x=6 

So

6^85 ≡ 6 mod 35


If any clarifications are needed or if you find any errors, please post.
5 0
3 years ago
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