To solve this problem,you need to use the formula d = rd (distance = rates x time)She runs at a speed of 7 mph and walks at a speed of 3 mph. Her distance running is d = 7trwhere tr is the time she spends running Her distance walking isd = 3twwhere tw is the time she spends walking The distances are the same so7tr = 3tw We also know that the total time is 4 hourstr + tw = 4tr = 4-tw Substitute this value of tr in the first equation7tr = 3tw7(4-tw) = 3tw28-7tw = 3tw28 = 10tw2.8 = tw Denise will spend 2.8 hours (2 hours, 48 minutes) walking back and 1.2 hours (1 hour, 12 minutes running.
Hope I helped :)
Compensation for the year:$51,810(deducting vehicle expense)
An employee makes 48,700+1,530(health insurance)+2,810(paid time)=
53,040. Also they receive 0.53 per mile x 9,000 miles=4,770
53,040 +4,770=57,810(employment compensation)
Personal work vehicle expense: 500 x 12 months=6,000
57,810 - 6,000= 51,810
Answer:3
Step-by-step explanation:
Select all the correct answers:
1) Yes
2) No
x=8→h(8)=2(8)^2+5(8)+2=2(64)+40+2=128+40+2→h(8)=170
x=8→f(8)=3^8+2=6,561+2→f(8)=5,563>170=h(8)
3) Yes
4) No
5) Yes
rg=[g(3)-g(2)]/(3-2)=[g(3)-g(2)]/1→rg=g(3)-g(2)
g(3)=20(3)+4=60+4→g(3)=64
g(2)=20(2)+4=40+4→g(2)=44
rg=64-44→rg=20
rf=f(3)-f(2)
f(3)=3^3+2=27+2→f(3)=29
f(2)=3^2+2=9+2→f(2)=11
rf=29-11→rf=18
rh=h(3)-h(2)
h(3)=2(3)^2+5(3)+2=2(9)+15+2=18+15+2→h(3)=35
h(2)=2(2)^2+5(2)+2=2(4)+10+2=8+10+2→h(2)=20
rh=35-20→rh=15
rg=20>18=rf
rg=20>15=rh
6) No
x=4→g(4)=20(4)+4=80+4→g(4)=84
x=4→h(4)=2(4)^2+5(4)+2=2(16)+20+2=32+20+2→h(4)=54
x=4→f(4)=3^4+2=81+2→f(4)=83>54=h(4)
f(4)=83<84=g(4)
Answer:
Valid Argument
Step-by-step explanation:
An argument is valid if its conclusion follows with certainty from its premises.
Given:
Premise: If it has an engine, I can fix it.
Premise: Cars have engines.
Conclusion: I can fix cars.
The argument is valid since the conclusion follows with certainty from the given premises.
