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julsineya [31]
3 years ago
5

Determine the solution to the system of equations {4x+15y=40

Mathematics
1 answer:
oee [108]3 years ago
8 0

Step-by-step explanation:

4x+15y= 40-5y= 20

= 4x+15y+5y= 40-20

4x+20y= 20

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satela [25.4K]

Answer:6+3w

Step-by-step explanation:

So, our number is "W"

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6+3w

Three times w=3w

Then, it says 6 more, so we add 6

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Answer:

Find the minimum or maximum value of the function g (I) = -3x^2 - 6x + 5. Describe the domain and range of the function, and where the function is increasing and decreasing. > -1 all real numbers The function The maximum value is I < 0 The domain is and the range is right of I left -1 is increasing to the of I= and decreasing to the 0 12 :: yo y0 :: 8 :: IS-1 :: -1 :: 0 :: I> 0 :: 1 :: 8 :: < 0 :: left :: all real numbers :: y -8 :: y < 8 :: y8​

Step-by-step explanation:

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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}&#10;\\\\=\frac{\sin{\frac{\pi}{3}}&#10;}{\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}}&#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)

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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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3 years ago
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alexandr1967 [171]

Answer:

3 minutes

Step-by-step explanation:

18 divided by 6 = 3 minutes

Hope this helps :)

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