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lesantik [10]
3 years ago
13

Last week joe drove his truck 4,563.8 miles. That brought the truck's total mileage to 99,999 miles. What was the truck's total

mileage before last week
Mathematics
1 answer:
adell [148]3 years ago
7 0

Answer:

95435.20 miles

Step-by-step explanation:

just add them both together

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Hans needs to rent a moving truck. suppose company a charges a rate of $40 per day and company B charges $60 fee plus $30 per da
kykrilka [37]
The time at which the cost for the services of the two companies would be in exactly 6 days. If we would compare the two charges, we would see that company A just charges 40 per day while B has a charge of 60, which is a standard fee, and a rate of 30 per day. So if we compute 40 x 6, we will have a total cost now of 240. If we compute 30 x 6, we will have a total cost now of 180 plus 60 which is the standard fee, and we will get 240. So in 6 days, the rates of the two companies would both accumulate to 240. 
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4 years ago
The estimated product of 20.7 and 9.18 after rounding both factors to the nearest whole number is
gogolik [260]
189 because 20.7 rounds to 21 and 9.18 rounds to 9
21 * 9 = 189

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3 years ago
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8 0
3 years ago
What value of b will cause the system to have an infinite number of solutions?
irga5000 [103]

b must be equal to -6  for infinitely many solutions for system of equations y = 6x + b and -3 x+\frac{1}{2} y=-3

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}

Let us bring the equations in same form for sake of simplicity in comparison

\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}

Now we have two equations  

\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}

Let us first see what is requirement for system of equations have an infinite number of solutions

If  a_{1} x+b_{1} y+c_{1}=0 and a_{2} x+b_{2} y+c_{2}=0 are two equation  

\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} then the given system of equation has no infinitely many solutions.

In our case,

\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}

 As for infinitely many solutions \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}

Hence b must be equal to -6 for infinitely many solutions for system of equations y = 6x + b and  -3 x+\frac{1}{2} y=-3

8 0
3 years ago
F(x)= 5x + 7. Evaluate F(5)
Vikentia [17]

Answer:

F(x) = 5x + 7

F(5) = 5*5 + 7 = 25 + 7 = 32

8 0
3 years ago
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