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3 years ago

Can someone please help me with these two questions pleasee

Mathematics

1 answer:

Nina [5.8K]3 years ago

7 0

Answer:

Step-by-step explanation:

The formula of an area of a triangle:

$A_\triangle=\dfrac{1}{2}ab\sin\theta$

a, b - adjacent sides

θ - included angle

=============================================

a = 9in, b = 14in, θ = 85°

sin85° ≈ 0.9962

Substitute:

$A_\triangle=\dfrac{1}{2}(9)(14)\sin85^o=\dfrac{1}{2}(126)(0.9962)\approx62.8 in^2$

a = 12 in, b = 5 in, θ = 135°

sin135° = sin(180° - 45°) = sin45° ≈ 0.7071 used sin(180° - x) = sin(x)

Substitute:

$A_\triangle=\dfrac{1}{2}(12)(5)\sin135^o=\dfrac{1}{2}(60)(0.7071)\approx21.2\ in^2$

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3 years ago

A veterinarian is enclosing a rectangular outdoor running area against his building for the dogs he cares for. He needs to maxim

NISA [10]

Answer:

a) The length of the building that should border the dog run to give the maximum area = 25feet

b) The maximum area of the dog run = 1250 s q feet²

Step-by-step explanation:

Step(i):-

Given function

A(x) = x (100-2x)

A (x) = 100x - 2x²...(i)

Differentiating equation (i) with respective to 'x'

$\frac{dA}{dx} = 100 (1) - 2 (2x)$

⇒ $\frac{dA}{dx} = 100 - 4 x$ ...(ii)

Equating zero

⇒ 100 - 4x =0

⇒ 100 = 4x

Dividing '4' on both sides , we get

x = 25

Step(ii):-

Again differentiating equation (ii) with respective to 'x' , we get

$\frac{d^{2} A}{dx^{2} } = -4 (1) < 0$

Therefore The maximum value at x = 25

The length of the building that should border the dog run to give the maximum area = 25

Step(iii)

Given A (x) = x ( 100 -2 x)

substitute 'x' = 25 feet

A(x) = 25 ( 100 - 2(25))

= 25(50)

= 1250

Conclusion:-

The maximum area of the dog run = 12 50 s q feet²

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4 years ago