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3 years ago

Help?

2 answers:

7 0

1 )
f ( - 10 ) = | - 10 + 6 | = | - 4 | = 4
Answer: D ) 4
2 )
f ( 1 ) = | 1 - 4 | + 5 = | - 3 | + 5 = 3 + 5 = 8
Answer: A ) 8
3 )
f ( - 6 ) = 8 - 3 * | - 6 + 2 | = 8 - 3 * | - 4 | = 8 - 3 / 4 = 8 - 12 = - 4
Answer: C ) - 4

Nikitich [7]3 years ago

3 0

What you need to do just plug the number into the x values in the function.

1. f(x) = x + 6
f(-10) = -10 + 6 = -4
The answer is C. -4

2. f(x) = (x − 4) + 5
f(1) = (1 - 4) + 5
= (-3) + 5 = 2
The answer is B. 2

3. f(x) = 8 − 3(x + 2)
f(−6) = 8 - 3(-6 + 2)
= 8 - 3(-4)
= 8 + 12 = 20
The answer is A. 20

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The estimated sum of the numbers fifty-eight (58) and twenty-nine (29) is equal to ninety (90).

<h3>What is the difference between a regular sum and an estimated sum?</h3>

In an estimated sum, the numbers are rounded; on the other hand, in a regular sum, the numbers are not changed.

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1. Round the numbers:

58 = 60 as this number is nearest to 60 than to 58.
29= 30 because of the same reason explained above.

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Compute the second partial derivatives ∂2f ∂x2 , ∂2f ∂x ∂y , ∂2f ∂y ∂x , ∂2f ∂y2 for the following function. f(x, y) = 2xy (x2 +

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Answer with step-by-step explanation:

We are given that a function

$f(x,y)=2xy(x^2+y^2)^2$

Differentiate partially w.r.t x

Then, we get

$\frac{\delta f}{\delta x}=2y(x^2+y^2)^2+8x^2y(x^2+y^2)=(x^2+y^2)(2x^2y+2y^3+8x^2y)=2(5x^2y+y^3)(x^2+y^2)$

Differentiate again w.r.t x

$\frac{\delta^2f}{\delta x^2}=2(10xy)(x^2+y^2)+4x(5x^2y+y^3)=20x^3y+20xy^3+20x^3y+4xy^3=40x^3y+24xy^3$

Differentiate function w.r.t y

$\frac{\delta f}{\delta y}=2x(x^2+y^2)^2+2xy\times 2(x^2+y^2)\times 2y$

$\frac{\delta f}{\delta y}=(x^2+y^2)(2x^3+2xy^2+8xy^2)=2(x^2+y^2)(x^3+5xy^2)$

Again differentiate w.r.t y

$\frac{\delta^2f}{\delta x^2}=2(2y)(x^3+5xy^2)+20xy(x^2+y^2)=4x^3y+20xy^3+20x^3y+20xy^3=24x^3y+40xy^3$

Differentiate partially w.r.t y

$\frac{\delta^2f}{\delta y\delta x}=2(2y(5x^2y+y^3)+(x^2+y^2)(5x^2+3y^2))=10x^4+36x^2y^2+10y^4$

$\frac{\delta^2f}{\delta y\delta x}=10x^4+36x^2y^2+10y^4$ $\frac{\delta^2f}{\delta x\delat y}=2(2x(x^3+5xy^2)+(3x^2+5y^2)(x^2+y^2))=10x^4+36x^2y^2+10y^4$

$\frac{\delta^2f}{\delta x\delat y}=10x^4+36x^2y^2+10y^4$

Hence, if f(x,y) is of class $C^2$ (is twice continuously differentiable), then the mixed partial derivatives are equal.

i.e $\frac{\delta^2f}{\delta y\delta x}=\frac{\delta^2f}{\delta x\delta y}$

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A) The domain of this relation would be 0,4,3,5, and 6. These are our possible x values for the function. Since this equation doesn't tell us to assume that it's a function then it's fine for us to list them separately.
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