A map has a scale of 1 in. : 38 ft use the given map distance to find the actual distance

Mathematics

1 answer:

Dvinal [7]2 years ago

8 0

Answer:

for q 19 its x=209ft and for 21 its 66.5 ft i forgot q 20

Step-by-step explanation:

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How do you solve?3(t+4)-2(2t+3)=-4

shepuryov [24]

3t+12-4t-6=-4
-t=-4+6-12
-t=-10
t=10
That's your answer.

8 0

2 years ago

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A new car is purchased for 17000 dollars. The value of the car depreciates at 12.25% per year. What will the value of the car be

Scilla [17]

Answer:

$P(t=14) = 17000 (1-0.1225)^{14}= 2728.404$

Step-by-step explanation:

for this case we have the following model for the cost of the car:

$P(t) = P_o (1-r)^t$

Where $P_o$ is the initial amount on this case 17000, t the amount of years after the initial year and r the depreciation rate on this case:

$r= \frac{12.25}{100}=0.1225$

And for t =14 we can replace into the equation and we got:

$P(t=14) = 17000 (1-0.1225)^{14}= 2728.404$

5 0

2 years ago

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

kondaur [170]

This is a problem of maxima and minima using derivative.

In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:

V = length x width x height

That is:

$V = xxy = x^{2}y$

We also know that the bin is constructed from 48 square feet of sheet metal, so:

Surface area of the square base = $x^{2}$

Surface area of the rectangular sides = $4xy$

Therefore, the total area of the cube is:

$A = 48 ft^{2} = x^{2} + 4xy$

Isolating the variable y in terms of x:

$y = \frac{48- x^{2} }{4x}$

Substituting this value in V:

$V = x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}$

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

$\frac{dv}{dx} = 48-3x^{2} =0$

Solving for x:

$x = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Solving for y:

$y = \frac{48- 4^{2} }{(4)(4)} = 2$

Then, the dimensions of the largest volume of such a bin is:

Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

$V = (4^{2} )(2) = 32 ft^{3}$