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3 years ago

A car can travel 234 miles on 26 gallons of gasoline. How far can it travel on 51 gallons?

Mathematics

2 answers:

castortr0y [4]3 years ago

7 0

Answer:

459 Miles

Step-by-step explanation:

fenix001 [56]3 years ago

5 0

If a car can travel 234 miles on 26 gallons, that means each gallon supplies 9 miles.
234/26=9
So, if a car has 51 gallons, it’ll be 51 x 9=459
The car can travel 459 miles on 51 gallons.

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Ronald measured a house and its lot and made a scale drawing. The front patio, which is 14 meters wide in real life, is 7 millim

zhannawk [14.2K]

Answer:

2 metre= 1millimetre

Step-by-step explanation:

From the measurement of the house The front patio, is 14 meters wide in real life, is 7 millimeters wide in the drawing

Real life= 14 meters

Drawing= 7 millimeters,

The scale he used is 2 metre= 1millimetre

Which means he magnify the drawing by scale of 2 metre to get the real life structure.

✓Let us clarify it

14 meters = 7 millimeters

2 metre= x millimeter

If we cross multiply

7 millimeters× 2 metre= 14 metre × X millimeters

Make X subject of the formula we have

X= 1millimetres

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3 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

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hram777 [196]

The answer is C hope this helped

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Step-by-step explanation:

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irina1246 [14]

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