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Yakvenalex [24]
3 years ago
13

A car uses 1 gallon of gasoline for every 25 miles driven. At the start of a trip the tank was full of gasoline. After driving 5

00 miles, there were 2 gallons of gasoline left in the tank. Which equation shows the amount of gasoline left in the tank, we, as a function of the number of miles driven, d
Mathematics
1 answer:
dezoksy [38]3 years ago
8 0

Answer:

d = 22 - x / 25

Step-by-step explanation:

The first thing is to calculate the tank capacity, we know that per gallon 25 miles are traveled and at 500 miles there were still two gallons, therefore:

500 miles / 25 miles / gallon = 20

and how there were still 2

20 + 2 = 22

the tank has a 22 gallon capacity

Therefore the equation would be:

d = 22 - x / 25

where x is the number of miles traveled.

for example, if they have traveled 500 miles it would be:

d = 22 - 500/25 = 22 - 20

d = 2 gallons

we can verify that it is correct because it is what the statement says

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drek231 [11]

graph B ( the lower one )

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graph A has more than 1 value of y for given values of x

for example ( - 1, 4 ) and  ( - 1, - 4 ) thus is not a function

graph B only has 1 value of y for a given value of x thus is a function




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3 years ago
Factor each expression.
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4 0
3 years ago
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State the number of possible triangles that can be formed using the given measurements.
romanna [79]

Answer:  39) 1              40) 2

                41) 1              42) 0

<u>Step-by-step explanation:</u>

39)     ∠A = ?        ∠B = ?       ∠C = 129°

            a = ?          b = 15         c = 45

Use Law of Sines to find ∠B:

\dfrac{\sin B}{b}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin B}{15}=\dfrac{\sin 129}{45}\rightarrow \quad \angle B=15^o\quad or \quad \angle B=165^o

If ∠B = 15°, then ∠A = 180° - (15° + 129°) = 36°

If ∠B = 165°, then ∠A = 180° - (165° + 129°) = -114°

Since ∠A cannot be negative then ∠B ≠ 165°

∠A = 36°        ∠B = 15°       ∠C = 129°       is the only valid solution.

40)      ∠A = 16°        ∠B = ?       ∠C = ?

             a = 15           b = ?         c = 19

Use Law of Sines to find ∠C:

\dfrac{\sin A}{a}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin 16}{15}=\dfrac{\sin C}{19}\rightarrow \quad \angle C=20^o\quad or \quad \angle C=160^o

If ∠C = 20°, then ∠B = 180° - (16° + 20°) = 144°

If ∠C = 160°, then ∠B = 180° - (16° + 160°) = 4°

Both result with ∠B as a positive number so both are valid solutions.

Solution 1:  ∠A = 16°        ∠B = 144°       ∠C = 20°    

Solution 2:  ∠A = 16°        ∠B = 4°       ∠C = 160°    

41)       ∠A = ?        ∠B = 75°       ∠C = ?

             a = 7           b = 30         c = ?

Use Law of Sines to find ∠A:

\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{7}=\dfrac{\sin 75}{30}\rightarrow \quad \angle A=13^o\quad or \quad \angle A=167^o

If ∠A = 13°, then ∠C = 180° - (13° + 75°) = 92°

If ∠A = 167°, then ∠C = 180° - (167° + 75°) = -62°

Since ∠C cannot be negative then ∠A ≠ 167°

∠A = 13°        ∠B = 75°       ∠C = 92°       is the only valid solution.

42)      ∠A = ?         ∠B = 119°       ∠C = ?

             a = 34         b = 34           c = ?

Use Law of Sines to find ∠A:

\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{34}=\dfrac{\sin 119}{34}\rightarrow \quad \angle A=61^o\quad or \quad \angle A=119^o

If ∠A = 61°, then ∠C = 180° - (61° + 119°) = 0°

If ∠A = 119°, then ∠C = 180° - (119° + 119°) = -58°

Since ∠C cannot be zero or negative then ∠A ≠ 61° and ∠A ≠ 119°

There are no valid solutions.

6 0
3 years ago
10y 5 +30y 3 −15y <br> Factor out the gcf then simplify the polynomial
frosja888 [35]

10y⁵ + 30y³ - 15y =

5y * 2y⁴ + 5y * 6y² + 5y * (-3) =

5y (2y⁴ + 6y² - 3)

7 0
2 years ago
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