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Elis [28]
3 years ago
12

Please Answer The Ones That Are Circled. (DON'T PUT DECIMALS FOR THE DIVISION PUT REMAINDER)if u can't see any tell me.

Mathematics
1 answer:
Alecsey [184]3 years ago
3 0
Lesson 2.7:
1) / divide 195 by 18 to get 10.833.
/ to get the whole number, round up to get '11'.
11 cartons


Lesson 2.8:
3) 
588 ÷ 84 = 7
5) 6,681 ÷ 53 = 126 with a remainder of 3

Lesson 2.9:
1) / let 'x' equal the inches of snow that reidville gets
/ '12x' equal the inches of snow that greenboro gets
/ you should set up the problem like: x + 12x = 65
/ add 'x + 12x' to get '13x'. 
/ divide both sides by '13'.
/ 5 goes into 13 to get 65. simplify the fraction to '5'.
/ x = 5
Reidville gets 5 inches of snow

2) / let 'x' equal the miles ansley ran in one month
/ you should set up the problem like: 15x + x = 192
/ add '15x + x' to get '16x'
/ divide both sides by '16'
/ 12 goes into 16 to get 192. simplify the fraction to 12.
/ x = 12
Ansley runs 180 miles (Zack only runs 12)



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so the investigator found the skid marks were 75 feet long hmmm what speed will that be?

s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ d=75 \end{cases}\implies s=\sqrt{30(0.7)(75)}\implies s\approx 39.69~\frac{m}{h}

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now, assuming Charlie was indeed going at 35 m/h, then his skid marks would have been

s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ s=35 \end{cases}\implies 35=\sqrt{30(0.7)d} \\\\\\ 35^2=30(0.7)d\implies \cfrac{35^2}{30(0.7)}=d\implies 58~ft\approx d

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