Question

Bantu dong no.1 smpe 10

Answer 1

1) f(x) = 6x + 1
f(-1) = 6(-1) + 1 = (-6)+ 1 = (-5)
f(5)=6×5=30
2)f(x)=3x²-4
f(-2)=3(-2)²-4=3×4-4=12-4=8
f(7)=3×7²-4=3×49-4=147-4=143
3)h(x)=x/3 + 5
h(x)=k
h(x)=k/3 + 5= k + 15
h(2n)=2n/3 + 5=2n + 15
4)p(n)=4n-5
15=4n-5~4n=15+5~4n=20~n=5
p(n)=19
19=4n-5~4n=19+5~4n=24~n=6
p(-n)=11
11=(-4n)-5~11+5=-4n~n=(-4)
5)f(x)=x³-7
-15=a³-7~a³=(-15)+7~a³=8~a=2
20=a³-7~a³=20+7~a³=27~a=3
1=(-a)³-7~-a³=8~a=-2
6)l(0)=1/2 x²-2=1/2 (0)²-2=-2
l(2)=1/2 (2)² - 2= 0

Answer 2

6b)x=3~f(3)=½(3)²-2=2,5
x=4~f(4)=½(4)²-2=6
x=5~f(5)=½(5)²-2=10,5
7)p(-2)=2(-2)-2=-6
p(2)=2(2)-2=2
b)p(x)=24~24=2x-2~2x=26~ x=13
8)f(p)=2p-17=1~2p=18~p=9
b)p²+2p-12=(9)²+2.9-12=97
9)f(2)=ax+b=6~2a+b=6
f(-5)=-5a+b=-1
-------------------- (-) eliminasi
jadi 7a=7~a=1
subtitusi ke 2a+b=6~b=6-2.1=4
b)f(-1)=-a+b=-11
f(3)=3a+b=1
---------------- (-) eliminasi
jadi -4a=-12~a=3
subtitusi ke 3a+b=1~b=1-3.3=-8
10)f(-3)=2/(-3a+b)=1
~-3a+b=2 (persamaan 1)
f(1)=2/a+b=⅓
~2.3=a+b~a+b=6 (persamaan 2)
persamaan saling di eliminasi
-3a+b=2
a+b=6
---------- (-)
-4a=-4~a=1
a+b=6~b=6-1=5
f(3)=2/(1.3+5)=2/8=¼
f(7)=2/(1.7+5)=2/12=1/6