Answer:
Step-by-step explanation:
Given the explicit function as
f(n) = 15n+4
The first term of the sequence is at when n= 1
f(1) = 15(1)+4
f(1) = 19
a = 19
Common difference d = f(2)-f(1)
f(2) = 15(2)+4
f(2) = 34
d = 34-19
d = 15
Sum of nth term of an AP = n/2{2a+(n-1)d}
S20 = 20/2{2(19)+(20-1)15)
S20 = 10(38+19(15))
S20 = 10(38+285)
S20 = 10(323)
S20 = 3230.
Sum of the 20th term is 3230
For the explicit function
f(n) = 4n+15
f(1) = 4(1)+15
f(1) = 19
a = 19
Common difference d = f(2)-f(1)
f(2) = 4(2)+15
f(2) = 23
d = 23-19
d = 4
Sum of nth term of an AP = n/2{2a+(n-1)d}
S20 = 20/2{2(19)+(20-1)4)
S20 = 10(38+19(4))
S20 = 10(38+76)
S20 = 10(114)
S20 = 1140
Sum of the 20th terms is 1140
Hello :
f(x) = (x − h)2 + k: a <span> vertex of( 2, 3)
h=2 and k=3</span>
By either long or synthetic division, it's easy to show that

The quartic will be exactly divisible by

when the numerator of the remainder term vanishes, or for those values of

such that

I'm not sure how to count the number of solutions (software tells me it should be 80), but hopefully this is a helpful push in the right direction.
The answer is -12
these are the steps on how to find that
-2x2y
-2(-3)2(y)
6 times -2
-12
<u>3x + 2y = 4</u>
Here are a few different ways
to represent the same line:
2y = 4 - 3x
y = -1.5x + 2
5y = -7.5x + 10
3x = 4 - 2y
x = -2/3y + 4/3