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blsea [12.9K]
2 years ago
8

I need this answers ASAP please. The mass of potatoes Lucan used is 5:2 the mass of carrots. The mass of potatoes is 9 kg more t

han the mass of carrots. Find the total mass of both ingredients. Please provide an explanation.
Mathematics
1 answer:
Gnoma [55]2 years ago
7 0

Answer:

Step-by-step explanation:

Known facts:

  • the ratio of mass potatoes to the mass of carrots is 5:2
  • the mass of potato is 9kg more than the mass of carrot

Let us set up the equation:

  • variables: mass of potatoes: p, the mass of carrots: c

p/c = 5/2 --> 2p = 5c --> 2p -5c = 0 -- equation 1

p = 9 + c --> p - c = 9 -- equation 2

Multiple equation 2 by 2

2p - 2c = 18 -- equation 3

subtract equation 1 and equation 3

-3c = -18

c = 6, so p = 15

So the total mass of both ingredients is 21 kg

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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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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}}
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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}

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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°

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Answer:

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5=\sqrt{\left(2-\left(-1\right)\right)^2+\left(k-1\right)^2}\\\\\mathrm{Square\:both\:sides}:\quad 25=k^2-2k+10\\25=k^2-2k+10\\\\\mathrm{Solve\:}\:25=k^2-2k+10:\\k^2-2k+10=25\\\\\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\k^2-2k+10-25=25-25\\k^2-2k-15=0\\\\\mathrm{Solve\:by\:factoring}\\\\\mathrm{Factor\:}k^2-2k-15:\quad \left(k+3\right)\left(k-5\right)\\\mathrm{Solve\:}\:k+3=0:\quad k=-3\\

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, list the first five terms of each sequence, and identify them as arithmetic or geometric.
Serhud [2]

Answer:

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The sequence is arithmetic

Step-by-step explanation:

Given that A(n + 1) = A(n) − 19 for n ≥ 1 and A(1) = −6

A(1) = -6

A(2) = A(1) − 19 = -6 - 19 = -25

A(3) = A(2) − 19 = -25 - 19 = -44

A(4) = A(3) − 19 = -44 - 19 = -63

A(5) = A(4) − 19 = -63 - 19 = -82

So the first 5 terms are -6,-25,-44,-63 and -82

If they are in arithmetic the common difference will be same, if geometric common ratio will be same.

Let us check common difference,

                       -25-(-6) = -19

                       -44-(-25) = -19

                       -63-(-44) = -19

                       -82-(-63) = -19

So common difference is same, so the sequence is arithmetic.

3 0
3 years ago
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