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aleksandr82 [10.1K]
3 years ago
7

If the perimeter of this triangle is to be no more than 77 feet, write an inequality and

Mathematics
1 answer:
Sergio039 [100]3 years ago
8 0

Given:

Consider the sides of the triangle are 4x+10, 2x-5 and 18 (in feet).

Perimeter of this triangle is to be no more than 77 feet.

To find:

The inequality and solution of the inequality.

Solution:

Perimeter of a triangle is the sum of all side.

Perimeter of given triangle =(4x+10)+(2x-5)+18

Perimeter of this triangle is to be no more than 77 feet. It means perimeter must be less than or equal to 77.

(4x+10)+(2x-5)+18\leq 77

(4x+2x)+(10-5+18)\leq 77

6x+23\leq 77

6x\leq 77-23

6x\leq 54

Divide both sides by 6.

x\leq \dfrac{54}{6}

x\leq 9

Therefore, the required inequality is (4x+10)+(2x-5)+18\leq 77 and the solution is x\leq 9.

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In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc
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Step-by-step explanation:

(a)

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Again consider,

\frac{b}{a}=\frac{\sin{B}}{\sin{A}}&#10;\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

Thus, the angle B is function of A is, B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

Now find \frac{dB}{dA}

Differentiate implicitly the function \sin{B}=\sqrt{\frac{3}{2}}\sin{A} with respect to A to get,

\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}

b)

When A=\frac{\pi}{4},B=\frac{\pi}{3}, the value of \frac{dB}{dA} is,

\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}

c)

In general, the linear approximation at x= a is,

f(x)=f'(x).(x-a)+f(a)

Here the function f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

At A=\frac{\pi}{4}

f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}

And,

f'(A)=\frac{dB}{dA}=\sqrt{3} from part b

Therefore, the linear approximation at A=\frac{\pi}{4} is,

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d)

Use part (c), when A=46°, B is approximately,

B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°

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