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Serjik [45]
4 years ago
5

Kimberly drives at a rate of 50 miles per hour on her trip across the country. If it takes Kimberly 200 gallons of gasoline to d

rive 500 miles, how many minutes will it take her to use x gallons of gasoline if she uses gasoline at the same average rate of consumption?
Mathematics
1 answer:
icang [17]4 years ago
6 0

Answer:

  3x minutes

Step-by-step explanation:

Kimberly desperately needs a more fuel-efficient vehicle. Her usage rate is ...

  (50 mi/h)(200 gal)/(500 mi) = 20 gal/h = 20 gal/(60 min) = 1/3 gal/min

Using gasoline at the rate of 1/3 gal/min, her time for x gallons is ...

  (x gal)/(1/3 gal/min) = 3x min

It will take 3x minutes for Kimberly to use x gallons at the same rate.

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I believe (3*8)*6=3*(8*6) illustrates the associative property of multiplication.
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24*6=3*48
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Which line is perpendicular to a line that has a slope of 1/2?​
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Twice a number is less than or equal to the sum of a number and 15
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Answer:itequalsjeh

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3 years ago
PLEASE HELP ASAP 35 PTS + BRAINLIEST TO RIGHT/BEST ANSWER
Semenov [28]

Answer:

(5,2,2)

Step-by-step explanation:

-3x+4y+2z = -3

2x-4y-z=0

y = 3x-13


Multiply the second equation by 2

2*(2x-4y-z)=0*2

4x -8y -2z =0

Add this to the first equation to eliminate z

-3x+4y+2z = -3

4x -8y -2z =0

-------------------------

x -4y = -3

Take the third equation and substitute it in for y

x - 4(3x-13) = -3

Distribute the 4

x - 12x +52 = -3

Combine like terms

-11x +52 = -3

Subtract 52 from each side

-11x +52-52 = -3-52

-11x = -55

Divide by -11

-11x/-11 = -55/-11

x=5

Now we can solve for y

y =3x-13

y =3*5 -13

y = 15-13

y=2

Now we need to find z

2x-4y-z=0

2(5) -4(2) -z=0

10-8 -z=0

2-z=0

Add z to each side

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2=z

x=5, y=2, z=2

(5,2,2)

3 0
3 years ago
Math be like 0-0????
pashok25 [27]

Answer:  A & C

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

HL is Hypotenuse-Leg

A) the hypotenuse from ΔABC ≡ the hypotenuse from ΔFGH

   a leg from ΔABC ≡ a leg from ΔFGH

   Therefore HL Congruency Theorem can be used to prove ΔABC ≡ ΔFGH

B) a leg from ΔABC ≡ a leg from ΔFGH

   the other leg from ΔABC ≡ the other leg from ΔFGH

   Therefore LL (not HL) Congruency Theorem can be used.

C) the hypotenuse from ΔABC ≡ the hypotenuse from ΔFGH

   at least one leg from ΔABC ≡ at least one leg from ΔFGH

   Therefore HL Congruency Theorem can be used to prove ΔABC ≡ ΔFGH

D) an angle from ΔABC ≡ an angle from ΔFGH

   the other angle from ΔABC ≡ the other angle from ΔFGH

   AA cannot be used for congruence.

5 0
3 years ago
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