1) f(x)=2x+6
f(2)=2(2)+6
=4+6
=10
2)f(x)=3x
f(a+1)=3(a+1)
=3a+3
3)f(x)=3x-1 and g(x)=5x+3
f(2)=3(2)-1
f(2)=6-1
=5
g(3)=5x+3
=5(3)+32
=15+3
=18
f(2)+f(3)=5+18
=23.
Answer:
(ab - 6)(2ab + 5)
Step-by-step explanation:
Assuming you require the expression factorised.
2a²b² - 7ab - 30
Consider the factors of the product of the coefficient of the a²b² term and the constant term which sum to give the coefficient of the ab- term
product = 2 × - 30 = - 60 and sum = - 7
The factors are - 12 and + 5
Use these factors to split the ab- term
= 2a²b² - 12ab + 5ab - 30 ( factor the first/second and third/fourth terms )
= 2ab(ab - 6) + 5(ab - 6) ← factor out (ab - 6) from each term
= (ab - 6)(2ab + 5) ← in factored form
Answer:
Step-by-step explanation:
Answer:
C. 272
Step-by-step explanation:
Since we know 85% of the students participated last year we can assume the same amount will participate this year so to find the answer you simply do the total number of students (320) multiplied by 85% or .85 to get the best guess of how many students will participate this year.
General equation of a circle with centre (h, k) is given by:
Now, the origin is the centre and radius is 20, so substituting these points in yields:
