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MissTica
4 years ago
10

Substitute the values of x and y into the expression -3x 2 + 2y 2 + 5xy - 2y + 5x 2 - 3y 2. Match that value to one of the numbe

rs.
1. x = 2, y = -1 A.-1
2. x = 0, y = 2.5 B.14
3. x = -1, y = -3 C.-11.25
4. x = 0.5, y = -1/10 D.9.17
5. x =3/3 , y =2/5 E.0.44
6. x = √2, y = √2 F. 1.665
Mathematics
1 answer:
Lynna [10]4 years ago
8 0
For x = 2 and y = -1

-3x²+2y²+5xy-2y+5x²-3y² ⇒ Simplify by collecting like terms
2x² - y² + 5xy -2y ⇒ then substitute the given value
2(2)² - (-1)² + 5(2)(-1) - 2(-1)
2(4) - (1) + 5(-2) + 2 
8 - 1 - 10 + 2
-1

Option 1 matches with option A
-----------------------------------------------------------------------------------------------------------

Given x = 0 and  y = 2.5
Substitute into 2x² - y² + 5xy -2y to get:

2(0)² - (2.5)² + 5(0)(2.5) - 2(2.5) = 0 - 6.25 + 0 - 4.5 = -11.25

Option 2 matches with option C
------------------------------------------------------------------------------------------------------------

Given x = -1 and y = -3
Substitute into 2x² - y² + 5xy - 2y, we obtain:
2(-1)² - (-3)² + 5(-1)(-3) - 2(-3) = 14

Option 3 matches with option B
-----------------------------------------------------------------------------------------------------------------

Given x = 0.5 and y = -1/10
Subsitute into 2x² - y² + 5xy - 2y, we obtain
2(0.5)² - (-1/10)² + 5(0.5)(-1/10) - 2(-1/10) = 0.44

Option 4 matches with option E
--------------------------------------------------------------------------------------------------------------

Given x = 3/4 and y = 2/5
Substitute into 2x² - y² + 5xy - 2y, we obtain
2(3/4)² - (2/5)² + 5(3/4)(2/5) - 2(2/5) = 1.665

Option 5 matches with option F

----------------------------------------------------------------------------------------------------------------

Given x = √2 and y= √2
Subsitute into 2x² - y² + 5xy - 2y, we obtain:
2(√2)² - (√2)² + 5(√2)(√2) - 2(√2) = 9.17

Option 6 matches with option D


                                           




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Step-by-step explanation:

(a)

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Now find \frac{dB}{dA}

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\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°

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