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3 years ago

8

what is the length of the diagonal of a non-regulation tennis court with length 20 feet and width 15 feet?

2 answers:

rodikova [14]3 years ago

8 0

Answer:

The diagonal of a non-regulation tennis court = 25 feet

Step-by-step explanation:

<u>Pythagorean theorem</u>

Hypotenuse² = Base² + Height²

The tennis court is like a rectangle.

We can consider the court as made of two right angled triangle

<u>To find the length of diagonal of court</u>

Here base = 15 feet and height = 20 feet

Diagonal or hypotenuse can be written as,

Diagonal ² = Base² + Height²

= 15² + 20²

= 225 + 400

= 625

Diagonal = √625 = 25 feet

Therefore the diagonal of a non-regulation tennis court = 25 feet

Semmy [17]3 years ago

6 0

Answer:

25 feet

Step-by-step explanation:

Basically that non-regulation tennis court is a rectangle. You want to know the length of the diagonal. If you draw it on paper, you'll see that this then become 2 triangles... of which you have 2 sides, and are seeking the hypotenuse. So....

H² = A² + B²

H² = 20² + 15² = 400 + 225 = 625

H = 25 feet.

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2 years ago

An aircraft manufacturer wants to determine the best selling price for a new airplane. The company estimates that the initial co

Blizzard [7]

Answer:

$(a)$ $C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2$

$p(x)=320-7.7p$

$R(x)=(320-7.7p)p=320p-7.7p^2$

$(b)$ $x=82 \text{planes}$

$(c)$ $p=\$30.91 M\;\; \text{per plane}$

$(d)$ maximum profit $=\$ 15.90M$

Step-by-step explanation:

Given that,

The company estimates that the initial cost of designing the aeroplane and setting up the factories in which to build it will be 500 million dollars.

The additional cost of manufacturing each plane can be modelled by the function.

$m(x)=20x-5x^{\frac{3}{4}}+0.01x^2$

$(a)$ Find the cost, demand (or price), and revenue functions.

$C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2$

$p(x)=320-7.7p$

$R(x)=(320-7.7p)p=320p-7.7p^2$

$(b)$ Find the production level that maximizes profit.

$f=R(x)-C(x)$

$\Rightarrow f=320p-7.7p^2-(500+20x-5x^{\frac{3}{4}}+0.01x^2)$

$\Rightarrow df=320dp-15.4pdp-20dx+5(\frac{3}{4} )x^{\frac{-1}{4} }dx-0.02xdx$

$x=320-7.7p$

$p=\frac{320-x}{7.7}$

$\frac{dp}{dx} = \frac{-1}{7.7}$

$\frac{df}{dx}=\frac{320}{-7.7} -\frac{15.4(320-x) }{7.7(\frac{-1}{7.7} )}-20+5\frac{3}{4} x^{\frac{-1}{4}} -0.02x=0$

$\Rightarrow -41.5584+83.1169-0.2597x-20+3.75x^{\frac{-1}{4} }-0.02x=0$

$\Rightarrow 21.5585+3.75x^{\frac{-1}{4} }-0.279x=0$

$\Rightarrow x=82 \text{planes}$

$(c)$ Find the associated selling price of the aircraft that maximizes profit.

$p=\frac{320-82}{7.7}$

$\Rightarrow p=\$30.91 M\;\; \text{per plane}$

$(d)$ Find the maximum profit.

Manufacturing cost of one plane is:

$m(1)=20-5+0.01$

$=\$15.01 M$

maximum profit $=\$(30.91-15.01)M$

$=\$15.90M$

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2 years ago

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mafiozo [28]

Answer:

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Step-by-step explanation:

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mezya [45]

First we need slope

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$\\ \sf\longmapsto m=\dfrac{12}{7}$

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$\\ \sf\longmapsto 12=12+b$

$\\ \sf\longmapsto b=12-12$

$\\ \sf\longmapsto b=0$

Now

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