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alekssr [168]
3 years ago
10

Y = 4r - 9 Complete the missing value in the solution to the equation. (3, )

Mathematics
2 answers:
Nostrana [21]3 years ago
5 0

Answer:

3

Step-by-step explanation:

Equation given = y=4r-9

substitute r=3

y=4*3-9

y=12-9

y=3

(3,3)

kozerog [31]3 years ago
4 0
The given equation: y=4r-9
Substitute the value of x=3 into the equation

y= 4(3) -9
y= 12-9 subtract
y=3

Answer (3,3)
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A truck that is 11 feet tall and 7 feet wide is traveling under an arch. The arch can be modeled by:
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Answer:

Yes

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Step-by-step explanation:

The equation of the arch is

y=-0.0625x^2+1.25x+5.75

Differentiating with respect to x we get

\dfrac{dy}{dx}=-2\times 0.0625x+1.25=-0.125x+1.25

Equating with zero

0=-0.125x+1.25\\\Rightarrow x=\dfrac{-1.25}{-0.125}\\\Rightarrow x=10

Double derivative of the equation

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So, the function is maximum at x=10

y=-0.0625x^2+1.25x+5.75=-0.0625\times 10^2+1.25\times 10+5.75\\\Rightarrow y=12

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So, the truck's center should be below the point (10,12) for maximum width to pass through.

Now truck is 11 feet tall so y=11

11=-0.0625x^2+1.25x+5.75\\\Rightarrow -0.0625x^2+1.25x-5.25=0\\\Rightarrow x=\frac{-1.25\pm \sqrt{1.25^2-4\left(-0.0625\right)\left(-5.25\right)}}{2\left(-0.0625\right)}\\\Rightarrow x=6,14

The maximum width of the truck can be 14-6=8\ \text{ft} if it is 11 feet tall.

So, the truck which has a width of 7 feet can fit under the arch.

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