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Setler [38]
2 years ago
8

Jacob drove a total of 15.235 miles over a 5-day period. How many miles did he drive each day if he drove the same distance each

day?
3.047 miles

3.047 miles

3.407 miles

3.407 miles

30.047 miles

30.047 miles

30.407 miles
Mathematics
2 answers:
8090 [49]2 years ago
6 0
15.235=5x
x= miles per day
your answer would be 3.047 miles
Virty [35]2 years ago
4 0

Answer:

3.047

Step-by-step explanation:

\frac{miles}{days} \\ \\\frac{15.235}{5} = \frac{?}{1} \\

When solving equations you multiply the denominator of one of the fractions by the others numerator. Here the available numerator is 15.235 and the others denominator is 1. This means that we do 15.235 times 1, which is equal to 15.235. The next step is to divide by the remaining denominator or numerator. Here the remaining denominator is 5. This means that we divide 15.235 by 5 which is equal to 3.074.

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F(5) = 5² +4*5=25+20 = 45
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f(5) + g(6)
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2 years ago
Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.
finlep [7]

Answer:

a) v = \frac{[L]}{[T]} = LT^{-1}

b) a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}

c) \int v dt = s(t) = [L]=L

d) \int a dt = v(t) = [L][T]^{-1}=LT^{-1}

e) \frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

v = \frac{ds}{dt}

a= \frac{dv}{dt}

Part a

If we do the dimensional analysis for v we got:

v = \frac{[L]}{[T]} = LT^{-1}

Part b

For the acceleration we can use the result obtained from part a and we got:

a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}

Part c

From definition if we do the integral of the velocity respect to t we got the position:

\int v dt = s(t)

And the dimensional analysis for the position is:

\int v dt = s(t) = [L]=L

Part d

The integral for the acceleration respect to the time is the velocity:

\int a dt = v(t)

And the dimensional analysis for the position is:

\int a dt = v(t) = [L][T]^{-1}=LT^{-1}

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

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