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vlabodo [156]
3 years ago
15

Anyone help plzzzzzzz?

Mathematics
1 answer:
Blizzard [7]3 years ago
7 0

The answer is D. The ones on the left are the domain. The ones on the right which the arrows point to are the range

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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
Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°

Use sine rule,

\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}

Again consider,

\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\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,

f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}

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°

8 0
3 years ago
Will mark brainliest! The triangles shown below must be congruent, true or false
olchik [2.2K]
True because they have the same size angles
8 0
2 years ago
Read 2 more answers
Dylan's expenditures for each of the past six months were: March $1,248.59; April, $1,365.38; May, $1,024.30; June, $1,100.40; J
ivann1987 [24]

Answer:

The average monthly expenditure is 1,155.35 $.

Step-by-step explanation:

The average of any sequency is given by the sum of it's individual parts divided by the number of parts it has. So in this case the average of monthly expenditure will be the sum of all individual expenditures divided by the number of months. The question can be solved like this:

average = (March+ April + May + June + July+ August)/6

average = (1,249.59 + 1,365.38 + 1,024.3 + 1,100.4 + 992,4 + 1,200.02)/6

average = 1,155.35 $

3 0
2 years ago
Read 2 more answers
What must be a factor of the polynomial function f(x) graphed on the coordinate plane below?
vesna_86 [32]

Answer:

x+2

Step-by-step explanation:

we know that

To find out the factors of a polynomial, determine the x-intercepts or zeros of the polynomial

Remember that

The x-intercepts of a polynomial are the values of x when the value of the polynomial is equal to zero

In this problem

Looking at the graph

For x=-2, f(x)=0

so

x=-2 is an x-intercept or zero of the polynomial

To find out the factor move the constant to the left side and equate to zero

x=-2

adds 2 both sides

x+2=-2+2\\x+2=0

therefore

The factor is (x+2)

5 0
3 years ago
Read 2 more answers
Derivative of cos(4x)
son4ous [18]
D/dx ( cos (4x)) = - sin (4x) · d/dx ( 4x) = - 4 sin (4x) 
7 0
3 years ago
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