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The functions f(x)=0.0875x-1.3x + 61.7 and g(x)=0.0875x +1.9x +11.1 model a car's stopping distance, f(x) or g(x), in feet,8001200-1000-pavement. The graphs of these functions are shown for {xx2 30) to the right. Use this information to complete parts (a) through600-traveling x miles per hour. Function f models stopping distance on dry pavement and function g models stopping distance on wet(d) below.400-(b)200-0-30 40 50 60 70 80 90 100a. Use the given functions to find the stopping distance on dry pavement and the stopping distance on wet pavement for a car traveling 35 miles per hourfeetThe stopping distance on dry pavement is(Round to the nearest whole number as needed)feetThe stopping distance on wet pavement is(Round to the nearest whole number as needed.)b. Based on your answers to part (a), which rectangular coordinate graph shows stopping distances on dry pavement and which shows stopping distances on wetpavement?Click to select your answer(s)WXto searchhpPACK ARDho12insprt scdeletehome$4&765Onumlockbackspace}RTUOhoGKLenterBpauseshiftaltctrl86LL

Question
The functions f(x)=0.0875x-1.3x + 61.7 and g(x)=0.0875x +1.9x +11.1 model a car's stopping distance, f(x) or g(x), in feet,
800
1200-
1000-
pavement. The graphs of these functions are shown for {xx2 30) to the right. Use this information to complete parts (a) through
600-
traveling x miles per hour. Function f models stopping distance on dry pavement and function g models stopping distance on wet
(d) below.
400-
(b)
200-
0-
30 40 50 60 70 80 90 100
a. Use the given functions to find the stopping distance on dry pavement and the stopping distance on wet pavement for a car traveling 35 miles per hour
feet
The stopping distance on dry pavement is
(Round to the nearest whole number as needed)
feet
The stopping distance on wet pavement is
(Round to the nearest whole number as needed.)
b. Based on your answers to part (a), which rectangular coordinate graph shows stopping distances on dry pavement and which shows stopping distances on wet
pavement?
Click to select your answer(s)
W
X
to search
hp
PACK ARD
ho
12
ins
prt sc
delete
home
$
4
&
7
6
5
O
num
lock
backspace
}
R
T
U
O
ho
G
K
L
enter
B
pause
shift
alt
ctrl
86
LL
help_outline

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The functions f(x)=0.0875x-1.3x + 61.7 and g(x)=0.0875x +1.9x +11.1 model a car's stopping distance, f(x) or g(x), in feet, 800 1200- 1000- pavement. The graphs of these functions are shown for {xx2 30) to the right. Use this information to complete parts (a) through 600- traveling x miles per hour. Function f models stopping distance on dry pavement and function g models stopping distance on wet (d) below. 400- (b) 200- 0- 30 40 50 60 70 80 90 100 a. Use the given functions to find the stopping distance on dry pavement and the stopping distance on wet pavement for a car traveling 35 miles per hour feet The stopping distance on dry pavement is (Round to the nearest whole number as needed) feet The stopping distance on wet pavement is (Round to the nearest whole number as needed.) b. Based on your answers to part (a), which rectangular coordinate graph shows stopping distances on dry pavement and which shows stopping distances on wet pavement? Click to select your answer(s) W X to search hp PACK ARD ho 12 ins prt sc delete home $ 4 & 7 6 5 O num lock backspace } R T U O ho G K L enter B pause shift alt ctrl 86 LL

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Step 1

(a) For x=35 we have to find f(35).

Answer: Stopping distance on dry pavement = 123 feet.

f(x)-0.0875х2-1.3х+61.7
f(35)-0.0875(35)-1.3(35)+61.7
f(35) 123.3875
f(35)123
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f(x)-0.0875х2-1.3х+61.7 f(35)-0.0875(35)-1.3(35)+61.7 f(35) 123.3875 f(35)123

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Step 2

For x=35 we have to find g(35).

Answer: Stopping ...

g(x)-0.0875x2+1.9x+11.1
g(35)-0.0875(35)2+1.9(35)+11.1
g(35)=184.7875
g(x)185
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g(x)-0.0875x2+1.9x+11.1 g(35)-0.0875(35)2+1.9(35)+11.1 g(35)=184.7875 g(x)185

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