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lana [24]
3 years ago
11

="TexFormula1" title="x^{2} +y^{2}\leq 1\\y-x^{2}\ \textgreater \ 0" alt="x^{2} +y^{2}\leq 1\\y-x^{2}\ \textgreater \ 0" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Ilya [14]3 years ago
7 0

Answer:

hi how are you

good afternoon

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156787 can go into 4 how many times???????
kiruha [24]

39196.75 is the answer to your problem

8 0
3 years ago
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
Y'all I am struggling Use the following functions to find each value below. f(x)=5x; g(x)=−2x+1; h(x)=x2+6x+8
Anni [7]

Answer:

see below the first three problems

Step-by-step explanation:

f(g(-2))

First, find g(-2) using function g(x). Then use that value as input for function f(x).

g(x) = -2x + 1

g(-2) = -2(-2) + 1

g(-2) = 5

f(x) = 5x

f(5) = 5(5)

f(5) = 25

f(g(-2)) = 25

g(h(3))

First, find h(3) using function h(x). Then use that value as input for function g(x).

h(x) = x^2 + 6x + 8

h(3) = 3^2 + 6(3) + 8 = 9 + 18 + 8

h(3) = 35

g(x) = -2x + 1

g(35) = -2(35) + 1 = -70 + 1

g(35) = -69

g(h(3)) = -69

f(g(3a))

First, find g(3a) using function g(x). Then use that value as input for function f(x).

g(x) = -2x + 1

g(3a) = -2(3a) + 1

g(3a) = -6a + 1

f(x) = 5x

f(-6a + 1) = 5(-6a + 1)

f(-6a + 1) = -30a + 5

f(g(3a)) = -30a + 5

5 0
3 years ago
To solve X/0.4 = 10, you would: Add 0.4 Subtract 0.4 Multiply 0.4 Divide by 0.4
Mrac [35]
Multiply by 0.4 on both sides to cancel the original 0.4 and change the 10
8 0
3 years ago
What is 45.3*4 equal please help me
xz_007 [3.2K]

Answer:

45.3 times 4 equals 181.2

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
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