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3 years ago

Dana is planning to drive 400 miles from

Mathematics

1 answer:

GaryK [48]3 years ago

5 0

Answer:

A. 8 gallons

B. Drive by car is $59.4 cheaper than travel by train

Step-by-step explanation:

According to the scenario, given data are as follows,

Total drive = 400 miles

Total drive per gallon = 50 miles

Fuel cost per gallon = $4.45

A. So, Total fuel required to drive 400 miles can be calculated as follows,

Total fuel required = Total drive ÷ Total drive per gallon

By putting value, we get,

Total fuel required = 400 ÷ 50

= 8 gallons

B. Total cost if drive by car = 8 gallons × $4.45 = $35.6

Cost if travel by train = $95

Hence it is clearly shows that drive by car is much cheaper than travel by train.

Cost saved by travel by car = $95 - $35.6 = $59.4

So, drive by car is $59.4 cheaper than travel by train.

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X=-6+y.x-3y=28. Iii fcugyvycuc

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Answer:

(-6+y)-3y=28

-2y-6=28

-2y=34

y=-17

Then,

x=-6-17

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Step-by-step explanation:

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3 years ago

Can anyone help me with this?0

sergeinik [125]

Answer:

C) (y - 1)(y - 5)

Step-by-step explanation:

Hi there!

First you separate your expression into groups.

$(y^2-5y)(-1y+5)$

Now we take the common multiple out of each.

$y(y - 5)-1(y - 5)$

Since we have two (y - 5)s, we can pull it out of the equation as its own common multiple.

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3 years ago

The length of a rectangle is 3 centimeters less than four times its width. Its area is 10 square centimeters. Find the dimension

Kay [80]

Answer:

W = 2 cm

L = 5 cm

Step-by-step explanation:

A rectangle is a four sided shape with 4 perpendicular angles. It has two pairs of parallel sides which are equal in distance: width and length. Its area, the amount of space inside it, can be found using the formula A = l*w. If the area is 10 cm² and the length is "3 cm less than 4 times the width" or 4w - 3, you can substitute and solve for w.

A = l*w

10 = (4w - 3)(w)

10 = 4w² - 3w

Subtract 10 from both sides to make the equation equal to 0. Then solve the quadratic by quadratic formula.

4w² - 3w - 10 = 0

Substitute a = 4, b = -3 and c = -10.

$w = \frac{3 +/- \sqrt{(-3)^2 - 4(4)(-10)} }{2(4)} = \frac{3 +/- \sqrt{9 +160)} }{8} = \frac{3 +/- \sqrt{169} }{8} = \frac{3+/-13}{8}$

There are two possible solutions which can be found.

3 + 13 / 8 = 16/ 8 = 2

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3 years ago

A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int

Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}$

By expanding the previous equation:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)$

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

$V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0$

Since there is no accumulation within the tank, expression is simplified to this:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}$

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

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$\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}$

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The instantaneous amount of salt in the tank is:

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