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Alexxx [7]
3 years ago
10

The gas tank of a 2013 Honda Accord holds 17.2 gallons. How many pounds of carbon dioxide are produced by burning an entire tank

of gas?
i don't know how but its a math problem
Mathematics
1 answer:
cricket20 [7]3 years ago
8 0
One gallon of gasoline (6.3 lbs.) produces about 20 lbs. of carbon dioxide. This may seem weird and untrue, however, most of that weight comes from the oxygen in the air. Following the math on this means that we multiply 17.2 gallons of gas by 20 lbs. of co2 per gallon of gas burned. 
This gives us the answer of 344 lbs. of carbon dioxide burned per tank of gas. (From a 2013 Honda Accord)
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a) square root of 11

Step-by-step explanation:

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4 years ago
Ebi, Jose, Derell, and Asami measured their heights. Ebi's height was 2.5 cm greater than Jose's height. Jose's height was 3.1 c
irga5000 [103]
Start with assigning each person with a variable to represent their height

Ebi: e
Jose: j
Derell: d
Asami: a

Ebi'd height was 2.5 cm greater than Jose's height

j + 2.5 = e

Jose's height was 3.1 cm greater than Derell's

d + 3.1 = j

Derell's height is 0.4 cm less than Asami's height

a - 0.4 = d

Ebi is 162.5 cm tall

e = 162.5

So, plug in 162.5 into any of the above equations were there is a variable of e

j + 2.5 = e

j + 2.5 = 162.5

Subtract 2.5 from both sides of the equation

j = 160 cm

Jose's height is 160 cm

Now, plug in 160 into any of the above equations where there is a j

d + 3.1 = j

d + 3.1 = 160

Subtract 3.1 from both sides of the equation 

d = 156.9 cm

Derell's height 156.9 cm

so, plug in 156.9 into any of the above equations where there is a d

a - 0.4 = d

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7 0
3 years ago
Read 2 more answers
Multiply (simplest form) <br> 6x^2/x(x-3)•(x-3)/10(x+2)
schepotkina [342]

Answer:

\frac{3x}{5(x+2)}

Step-by-step explanation:

Rewrite the division as a fraction.

8 0
3 years ago
Simplify the expression using trigonometric identities: sec (–θ) – cos θ.
ololo11 [35]

Answer:

sin θ . tan θ

Step-by-step explanation:

Note : -

sec ( - θ ) = sec θ

Formula / Identity : -

sec θ = 1 / cos θ

sec ( - θ ) - cos θ

= [ 1 / cos θ ] - cos θ

{ LCM = cos θ }

= [ 1 / cos θ ] - [ cos²θ / cos θ ]

= [ 1 - cos²θ ] / cos θ

{ 1 - cos²θ = sin²θ }

= sin²θ / cos θ

{ sin²θ = sin θ . sin θ }

= sin θ . sin θ / cos θ

{ sin θ / cos θ = tan θ }

= sin θ . tan θ

Hence, simplified.

4 0
2 years ago
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