An example for #1 would be that: 15 is divisible by 3, but not 9. every third multiple of 3, (9, 18, 27,...) is divisible by 9 because 9 is three times the size of 3.
For part two the first number that I thought of is 23.
An expression is a mathematical phrase containing at least one variable :)
Step-by-step explanation:
8x-5= 6x + 1
8x - 6x = 1+5
2x = 6
x= 2/6 = 1/3
x= 0.333
The answers are complex numbers...
Here's why...
---------
p+q=18
pq=82
---------
Therefore:
p=18-q
p=82/q
------------
Therefore:
18-q=82/q
q(18-q)=q(82/q)
18q-q²=82
(-1)(18q-q²)=82(-1)
q²-18q=-82
(q-9)²-9²=-82
(q-9)²-81=-82
(q-9)²=-82+81
(q-9)²=-1
q-9=-√(-1)=-i
q-9=√(-1)=i
-----------
Therefore:
q=9-i
q=9+i
-------------
ANSWERS:
When q=9-i, p=9+i
When q=9+i, p=9-i
-----------
Proof:
p+(9-i)=18
p+9-i=18
p=18-9+i
p=9+i
....
p+(9+i)=18
p+9+i=18
p=18-9-i
p=9-i
--------------
More proofs:
p=82/(9+i)
p=(82/(9+i))*((9-i)/(9-i))
p=(82(9-i))/(81-9i+9i-i²)
p=(82(9-i))/(81-(-1))
p=(82(9-i))/82
p=9-i
---------------
p=82/(9-i)
p=(82/(9-i))*((9+i)/(9+i))
p=(82(9+i))/(81+9i-9i-i²)
p=(82(9+i))/(81-(-1))
p=(82(9+i))/82
p=9+i
<h3>Given</h3>
right triangle with side "a" = 12 and hypotenuse "c" = 15
<h3>Find</h3>
cos(∠A)
<h3>Solution</h3>
The Pythagorean theorem tells you
... b = √(c² - a²) = √(225 - 144) = √81 = 9
The definition of the cosine tells you
... cos(A) = b/c = 9/15
... cos(A) = 3/5 . . . . . . reduce the fraction
