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3 years ago

Is this equation linear or nonlinear y=4x(8-x)

Mathematics

2 answers:

sweet-ann [11.9K]3 years ago

5 0

It is---------- linear--------------

FromTheMoon [43]3 years ago

5 0

Y=4x(8-x) distributes to y=32x-4x^2
The way it distributes as y=-4x^2+32x has a square power so it is quadratic not linear

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Step-by-step explanation:

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3 years ago

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3 years ago

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3 years ago

The combustion of one gallon of automobile fuel produces about 5 pounds of carbon (in CO₂). Two autos are making a trip of 600 m

rewona [7]

Answer:

Approximately 50 pounds less carbon (in CO₂) will be produced by the second auto on this trip.

Step-by-step explanation:

We have been given that the combustion of one gallon of automobile fuel produces about 5 pounds of carbon (in CO₂). Two autos are making a trip of 600 miles.

Let us find amount of fuel used by autos by dividing the total distance by their mileage.

$\text{Fuel used by 1st auto}=\frac{\text{ 600 miles}}{\frac{\text{ 20 miles}}{\text{gallon}}}$

$\text{Fuel used by 1st auto}=\frac{\text{ 600 miles}}{\text{ 20 miles}}\times \text{gallon}$

$\text{Fuel used by 1st auto}=30 \text{ gallons}$

$\text{Fuel used by 2nd auto}=\frac{\text{ 600 miles}}{\frac{\text{ 30 miles}}{\text{gallon}}}$

$\text{Fuel used by 2nd auto}=\frac{\text{ 600 miles}}{\text{ 30 miles}}\times \text{gallon}$

$\text{Fuel used by 2nd auto}=20 \text{ gallons}$

Let us find the difference of fuel used by both autos.

$\text{Number of gallons that 2nd auto used less than 1st auto}=30-20$

$\text{Number of gallons that 2nd auto used less than 1st auto}=10$

Since one gallon of automobile fuel produces about 5 pounds of carbon, so 10 gallons of fuel will produces $5\times 10=50$ pounds of carbon.

Since 2nd auto used 10 gallons less than 1st auto, therefore, 2nd auto will produce 50 pounds less carbon than 1st auto.

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3 years ago

Use the Pythagorean Theorem to find the missing length in the right triangle

Gelneren [198K]

Answer:

it depends

Step-by-step explanation:

If 17 is the hypotenuse

$c=17\\\\b=12\\\\a=?\\\\c^{2} =a^2+b^2\\\\17^2=a^2+12^2\\\\17^2-12^2=a^2\\\\(17-12)(17+12)=a^2\\\\5(27)=a^2\\\\a=\sqrt{9*3*5} \\\\a=3\sqrt{15} \\\\a=11.6$

if you are looking for the hypotenuse

$a=17\\\\b=12\\\\c=?\\\\a^2+b^2=c^2\\\\17^2+12^2=c^2\\\\289+144=c^2\\\\433=c^2\\\\c=\sqrt{433} \\\\c=20.8$

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3 years ago