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Fed [463]
3 years ago
10

the running back for the Bulldogs football team carried the ball nine times for a total loss of 15 3/4 yards find the average ch

ange in the field position on each run
Mathematics
1 answer:
Vitek1552 [10]3 years ago
6 0
The average change in the field position on each run is. -1 3/4
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X − 3 ≥ 6<br><br> pls help me
Fantom [35]

Answer:

x \geq 9

Step-by-step explanation: i took a quiz with this on it

6 0
3 years ago
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Karen Sullivan delivers newspapers for the Tribune dispatch. She receives 18.2 cents per paper, six days a week (for the daily p
mario62 [17]

Answer:

About $241.11

Step-by-step explanation:

So, Karen receives 18.2 cents per paper.

She delivers 124 paper per day.

In other words, on days other than Sunday, she will make a total of:

18.2(124)=2256.8\text{ cents}

On Sunday, each paper is sold for $0.70 or 70 cents. She also sells 151 Sunday papers. Thus, on a Sunday, she will make a total of:

70(151)=10570\text{ cents}

Therefore, in one week, she will do the first equation six times and the Sunday equation once. Thus, her total pay will be:

6(2256.8)+10570=24110.8\text{ cents}\approx\$241.11

4 0
3 years ago
Mixed 3 liters 20% solutions with 2 liters 70% solution. What’s the final concentration,
Advocard [28]

Answer:

(3×20%+2×70%)/3+2=40%

Step-by-step explanation:

Assuming the potions are the same type or do mix then probably the concentration of the potion depends on the type of reaction they have to each other.

Yet we can average the percentage of the active ingredient by the principle mentioned above

5 0
3 years ago
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Find the number of sides of a regular polygon in which the measure of an interior angle is three times the measure of an exterio
Serhud [2]
The sum of the interior angles of any polygon is given by (n-2)180°, where n is the number of sides.

Also sum of all the exterior angles of any polygon is 360°

Given (n-2)180° = 3(360°)

=> n-2 = 6

=> n = 8.

Hence the number of sides of a polygon is 8
6 0
2 years ago
Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g
AysviL [449]

Answer:

The quantity of salt at time t is m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} }), where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}

Where:

c - The salt concentration in the tank, as well at the exit of the tank, measured in \frac{pd}{gal}.

\frac{dc}{dt} - Concentration rate of change in the tank, measured in \frac{pd}{min}.

V - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

V \cdot \frac{dc}{dt} + 6\cdot c = 3

60\cdot \frac{dc}{dt}  + 6\cdot c = 3

\frac{dc}{dt} + \frac{1}{10}\cdot c = 3

This equation is solved as follows:

e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }

\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }

e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt

e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C

c = 30 + C\cdot e^{-\frac{t}{10} }

The initial concentration in the tank is:

c_{o} = \frac{10\,pd}{60\,gal}

c_{o} = 0.167\,\frac{pd}{gal}

Now, the integration constant is:

0.167 = 30 + C

C = -29.833

The solution of the differential equation is:

c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }

Now, the quantity of salt at time t is:

m_{salt} = V_{tank}\cdot c(t)

m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })

Where t is measured in minutes.

7 0
3 years ago
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