Use the Pythagorean Theorem to find the m 14.1 X X= 14.7

Find expressions for the partial derivatives of the following functions:

nlexa [21]

Answer:

Step-by-step explanation: partial derivative is the differentiation of one variable e.g. X while leaving the values of the other variable e.g. Y

These four questions A, B, C and D have different functions separated by commas. I will not assume the commas to be something else like a plus sign.

A. f(x) = g'(x).k(y) , g'(x) + h(y)

f(y) = k'(y).g(x) , g(x) + h'(y)

B. f(x) = g'x (x+y)

f(y) = g'y (x+y) , h'y (y+z)

f(z) = h'z (y+z)

C. f(x) = f'x (xy) , f'x (zx)

f(y) = f'y (xy) , f'y (yz)

f(z) = f'z (yz) , f'z (zx)

D. f(x) = f'x (x) , g'(x) , h'x (x,y)

f(y) = h'x (x,y)

These are the partial derivative expressions for each variable in each function. You will need to pay a lot of attention to understand:

* while differentiating X alone, functions in Y which are separated by commas from the functions in X, are ignored totally because they are different questions

* In functions where X added to Y is in a bracket e.g. (x+y), to find the derivative of X, Y isn't thrown away because they are joined (by a plus sign) the derivative of X alone in this case would be f'x (x+y)

* f(x), just like g(x), simply means/represents a function in X hence f'(x) means the differentiation of all X-terms in that function

6 0

2 years ago

6th grade math Help plz

irina1246 [14]

How do you say that having any confusion he said wouldn’t be able to consist of all the new stages

8 0

2 years ago

What value ofx is in the solution set of 3(x-4)>5x +2?

kvv77 [185]

$\boldsymbol{\sf{3(x-4) > 5x +2 }}$

Use the distributive property to multiply 3 by x−4.

$\boldsymbol{\sf{3x-12 > 5x+2 }}$

Subtract 5x on both sides.

$\boldsymbol{\sf{3x-12-5x > 2 } }$

Combine 3x and −5x to get −2x.

$\boldsymbol{\sf{-2x-12 > 2 }}$

Add 12 to both sides.

$\boldsymbol{\sf{-2x > 2+12 \ \longmapsto \ \ [Add \ 2 +12]}}$

$\boldsymbol{\sf{-2x > 14}}$

Divide both sides by −2. Since −2 is < 0, the inequality direction is changed.

$\boldsymbol{\sf{x < \dfrac{14}{-2} \ \ \longmapsto \ \ \ [Split]}}$

$\boxed{\boldsymbol{\sf{x < -7 }}}$

So in the inequality 3(x-4)>5x +2, the value of x is -7.

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8 0

1 year ago

16z+29z=pz-v solve for z

Margarita [4]

16z + 29z=pz - v

45z - pz = -v

z(45 - p) = -v

z = $\frac{-v}{(45 - p)}$

8 0

3 years ago

Read 2 more answers

Graphing polynomial functions?

Leni [432]

NOTES:

Degree: the largest exponent in the polynomial

End Behavior:

Coefficient is POSITIVE, then right side goes to POSITIVE infinity
Coefficient is NEGATIVE, then right side goes to NEGATIVE infinity
Degree is EVEN, then left side is SAME direction as right side
Degree is ODD, then left side is OPPOSITE direction as right side

Multiplicity (M): the exponent of the zero. e.g. (x - 3)² has a multiplicity of 2

Relative max/min: the y-value of the vertices.

Find the axis of symmetry (the midpoint of two neighboring zeros)
Plug the x-value from 1 (above) into the given equation to find the y-value. (which is the max/min)
Repeat 1 and 2 (above) for each pair of neighboring zeros.

Rate of Change: slope between the two given points.

********************************************************************************************

1. f(x) = (x-1)²(x + 6)

a) Degree = 3

b) end behavior:

Coefficient is positive so right side goes to positive infinity
Degree is odd so left side goes to negative infinity

c) (x - 1)²(x + 6) = 0

x - 1 = 0 x + 6 = 0

x = 1 (M=2) x = -6 (M=1)

d) The midpoint between 1 and -6 is -3.5, so axis of symmetry is at x = -3.5

y = (-3.5 - 1)²(-3.5 + 6)

= (-4.5)²(2.5)

= 50.625

50.625 is the relative max

e) see attachment #1

f) The interval at which the graph increases is: (-∞, -3.5)U(1, ∞)

g) The interval at which the graph decreases is: (-3.5, 1)

h) f(-1) = (-1 - 1)²(-1 + 6)

= (-2)²(5)

= 20

f(0) = (0 - 1)²(0 + 6)

= (-1)²(6)

= 6

Find the slope between (-1, 20) and (0, 6)

m = $\frac{20-6}{-1-0}$

= $\frac{14}{-1}$

= -14

********************************************************************************************

2. y = x³+3x²-10x

= x(x² + 3x - 10)

= x(x + 5)(x - 2)

a) Degree = 3

b) end behavior:

Coefficient is positive so right side goes to positive infinity

Degree is odd so left side goes to negative infinity

c) x(x + 5)(x - 2) = 0

x = 0 x + 5 = 0 x - 2 = 0

x = 0 (M=1) x = -5 (M=1) x = 2 (M=1)

d) The midpoint between -5 and 0 is -2.5, so axis of symmetry is at x = -2.5

y = -2.5(-2.5 + 5)(-2.5 - 2)

= -2.5(2.5)(-4.5)

= 28.125

28.125 is the relative max

The midpoint between 0 and 2 is 1, so axis of symmetry is at x = 1

y = 1(1 + 5)(1 - 2)

= 1(6)(-1)

= -6

-6 is the relative min

e) see attachment #2

f) The interval at which the graph increases is: (-∞, -2.5)U(1, ∞)

g) The interval at which the graph decreases is: (-2.5, 1)

h) f(-1) = -1(-1 + 5)(-1 - 2)

********************************************************************************************

3. y = -x(x + 2)(x - 7)(x - 3)

a) Degree = 4

b) end behavior:

Coefficient is negative so right side goes to negative infinity

Degree is even so left side goes to negative infinity

c) -x(x + 2)(x - 7)(x - 3) = 0

-x = 0 x + 2 = 0 x - 7 = 0 x - 3 = 0

x = 0 (M=1) x = -2 (M=1) x = 7 (M=1) x = 3 (M=1)

d) The midpoint between -2 and 0 is -1, so axis of symmetry is at x = -1

y = -(-1)(-1 + 2)(-1 - 7)(-1 - 3)

= 1(1)(-8)(-4)

= 32

32 is a relative max

The midpoint between 0 and 3 is 1.5, so axis of symmetry is at x = 1.5

y = -(1.5)(1.5 + 2)(1.5 - 7)(1.5 - 3)

= -1.5(3.5)(-5.5)(-1.5)

= -43.3125

-43.3125 is the relative min

The midpoint between 3 and 7 is 5, so axis of symmetry is at x = 5

y = -(5)(5 + 2)(5 - 7)(5 - 3)

= -5(7)(-2)(2)

= 140

140 is the relative max

e) see attachment #3

f) The interval at which the graph increases is: (-∞, -1)U(1.5, 5)

g) The interval at which the graph decreases is: (-1, 1.5)U(5, ∞)

h) f(-1) = -(-1)(-1 + 2)(-1 - 7)(-1 - 3)

= 1(1)(-8)(-4)

= 32

f(0) = -(0)(0 + 2)(0 - 7)(0 - 3)

= 0

Find the slope between (-1, 32) and (0, 0)

m = $\frac{32-0}{-1-0}$

= $\frac{32}{-1}$

= -32

5 0

3 years ago