3(x−5)/2=2(x+5)/3
Cross-multiply.
(3(x−5))/2=(2(x+5))/3
3(x−5)*(3)=2(x+5)*(2)
9x−45=4x+20
Subtract 4x from both sides.
9x−45−4x=4x+20−4x
5x−45=20
Add 45 to both sides.
5x−45+45=20+45
5x=65
Divide both sides by 5.
5x/5=65/5
x=13
Basically the remainder theorem links the remainder of division by a binomial with the value of a function at a point while the factor theorem links the factors of polynomial to its zeros
The answer for #4 is C. 21x-82
Let a,b,c,d and e are five numbers
mean = (a +b+ c+ d+e)/5 = 30
⇒ (a +b+ c+ d+e) = 150_______(1)
now, Let the number excluded be a
then, new mean = (b+ c+ d+e)/4 = 28
⇒ (b+ c+ d+e)= 112
putting this value in (1),
⇒a + 112 = 150
⇒a = 150 -112 = 38
excluded number = 38
