Hello,
f(x)=2x^3+4x²-32x+18
f(3)=2*3^3+4*3²-32*3+18=12
Remainder =12
Just work this problem out backwards. First write out an equation.
(x+4)•3=30
Now solve :
(x+4)•3=30
3x+12=30
-12 -12
3x=18
/3 /3
x=6
Your answer is 6!
Hoped I helped!
Answer:
No
Step-by-step explanation:
2x+5.2 + 3x+ 1.2 +2x+6.2= 180
7x+ 12.6=180
7x 12.6 - 12.6=180 -12.6
7x = 167.4
x = 23.914
should equal 60
Answer:
Possible locker numbers for Selma are: 3, 5 and 29
Step-by-step explanation:
First of all, let us have a look at the definition of a prime number.
A <em>prime number</em> is a number which is divisible either by 1 or the number itself.
No other number divides the prime number other than 1 and the number itself.
Now, let us factorize the given number 435 and let us observe the what all possibilities are there.

Factorizing 145 further:

No further factors are possible.
Therefore 3, 5 and 29 are the prime factors of 435.
So, the answer is:
Possible locker numbers for Selma are: 3, 5 and 29