A relation is a function if it has only One y-value for each x-value. Functions f(2/3)=2/9 for f(x)=2x²-4x+2 and f(1/4)==-21/4 for f(x)=4x²+2x-6
<h3>What is a function?</h3>
A relation is a function if it has only One y-value for each x-value.
The given function is
f(x)=2x²-4x+2
Put x=2/3
f(2/3)=2(2/3)²-4(2/3)+2
=2(4/9)-8/3+2
=8/9-8/3+2
=(8-24+18)/9
f(2/3)=2/9
Now f(x)=4x²+2x-6
Put x=1/4
f(1/4)=4(1/4)²+2(1/4)-6
=4/16+2/4-6
=1/4+1/2-6
= 1+2-24/4
f(1/4)==-21/4
Hence functions f(2/3)=2/9 for f(x)=2x²-4x+2 and f(1/4)==-21/4 for f(x)=4x²+2x-6
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Answer:
Answer:
2x • (x2 - 2xy + 5x - 10y)
Step-by-step explanation:
Step 1 :
Equation at the end of step 1 :
(((2•(x3))+(10•(x2)))-(22x2•y))-20xy
Step 2 :
Equation at the end of step 2 :
(((2 • (x3)) + (2•5x2)) - 22x2y) - 20xy
Step 3 :
Equation at the end of step 3 :
((2x3 + (2•5x2)) - 22x2y) - 20xy
Step 4 :
Step 5 :
Pulling out like terms :
5.1 Pull out like factors :
2x3 - 4x2y + 10x2 - 20xy =
2x • (x2 - 2xy + 5x - 10y)
Final result :
2x • (x2 - 2xy + 5x - 10y)
Step-by-step explanation:
Answer:
A) 10,560 ft. B) 32 more laps.
Step-by-step explanation:
A) Four laps around a track is equal to one mile, and there are 5,280 feet in one mile. If Roger ran eight laps, he ran two miles. Therefore: 5,280 ft. times 2, is equal to 10,560 ft.
B) We know that Roger ran two miles Monday. We also know that four laps around a track is equal to one mile. In order to run 10 miles, he would have to run 40 laps, and since he already ran 8, we can subtract that from 40. You are left with 32 more laps.
Step-by-step explanation:
a little messy but hope it helps :)) !!
