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3 years ago

Solve for w. 4. P = 2L + 2w

Mathematics

1 answer:

Maksim231197 [3]3 years ago

3 0

Answer: $W=\frac{P-2L}{2}$

Step-by-step explanation:

$P=2L+2W$

Isolate 2W by substracting 2L on both sides.

$-2L+P=2L+2W-2L$

$-2L+P=2W$

Isolate W by dividing by 2.

$\frac{-2L+P}{2} =\frac{2W}{2}$

$\frac{-2L+P}{2}=W$ or $W=\frac{P-2L}{2}$

It's the same.

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The cable between two towers of a suspension bridge can be modeled by the function shown, where x and y are measured in feet. Th

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a) x = 200 ft

b) y = 50 ft

Domain: $0\leq x \leq 400$

Range: $50\leq y \leq 150$

Step-by-step explanation:

The function that models the carble between the two towers is:

$y=\frac{1}{400}x^2-x+150$

where x and y are measured in feet.

Here we want to find the lowest point of the cable: this is equivalent to find the minimum of the function $y(x)$ .

In order to find the minimum of the function, we have to calculate its first derivative and require it to be zero, so:

$y'(x)=0$

The derivative of y(x) is:

$y'(x)=\frac{1}{400}\cdot 2 x^{2-1}-1=\frac{1}{200}x-1$

And requiring it to be zero,

$\frac{1}{200}x-1=0$

Solving for x,

$\frac{1}{200}x=1\\x=200 ft$

In order to find how high is the road above the water, we have to find the value of y (the height of the cable) at its minimum value, because that is the point where the cable has the same height above the water as the road.

From part a), we found that the lowest position of the cable is at

$x=200 ft$

If we now substitute this value into the expression that gives the height of the cable,

$y=\frac{1}{400}x^2-x+150$

We can find the lowest height of the cable above the water:

$y=\frac{1}{400}(200)^2-200+150=50 ft$

Therefore, the height of the road above the water is 50 feet.

Domain:

The domain of a function is the set of all possible values that the independent variable x can take.

In this problem, the extreme points of the domain of this function are represented by the position of the two towers.

From part a), we calculated that the lowest point of the cable is at x = 200 ft, and this point is equidistant from both towers. If we set the position of the tower on the left at

$x=0 ft$