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3 years ago

Q4.A/Find the linearization L(x, y) of the function f(x, y) = (x + y + 2)² at p. = (1,2)

Mathematics

1 answer:

Dafna11 [192]3 years ago

3 0

Answer:

Find the linearization L(x,y) of the function at each point. f(x,y) = x2 + y2 + 1 a. (4,0) b. (2,0) a. L(x,y) = Find the linearization L(x,y,z) of the function f(x,y,z) = 1x2 + y2 +z2 at the points (7,0,0), (3,4,0), and (4,4,7). The linearization of f(x,y,z) at (7,0,0) is L(x,y,z)= (Type an exact answer, using radicals as needed.)

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Find the particular solution that satifies the differential equation and the initial condition. f"(x) = x2 f'(0) = 8, f(0) = 4

viva [34]

Answer:

f(x) = x^4/12 + 8x + 4

Step-by-step explanation:

f"(x) = x^2

Integrate both sides with respect to x

f'(x) = ∫ x^2 dx

= (x^2+1)/2+1

= (x^3)/3 + C

f(0) = 8

Put X = 0

f'(0) = 0+ C

8 = 0 + C

C= 8

f'(x) = x^3/3 + 8

Integrate f(x) again with respect to x

f(x) = ∫ (x^3 / 3 ) +8 dx

= ∫ x^3 / 3 dx + ∫8dx

= x^(3+1) / 3(3+1) + 8x + D

= x^4/12 + 8x + D

f(0) = 4

Put X = 0

f(0) = 0 + 0 + D

4 = D

Therefore

f(x) = x^4 /12 + 8x + 4

5 0

3 years ago

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Alisiya [41]

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2 years ago

What is the x-and y-intercept of the graph 4x -5y =40??

umka21 [38]

Answer:

x-intercept = 10

y-intercept = -8

Step-by-step explanation:

$4x-5y=40$

To identify the y-intercept, we will need to format the equation in the form $y=mx+b$ . We can start by subtracting $4x$ from both sides of the equation:

$-5y=-4x+40$

Divide both sides of the equation by the coefficient of $y$ , which is $-5$ :

$y=\frac{-4}{-5} x - 8$

The $b$ is representative of the y-intercept in $y=mx+b$ .

Identify $b$ in the equation $y=\frac{-4}{-5} x - 8$ :

$b=-8$

Therefore, our y-intercept is -8.

To solve for the x-intercept, replace the $y$ in the equation $y=\frac{-4}{-5} x - 8$ with a zero:

$0=\frac{-4}{-5} x - 8$

Add $8$ to both sides of the equation:

$8=\frac{-4}{-5} x$

You can go ahead and divide $-4$ by $-5$ to get rid of the fraction:

$8=0.8x$

Divide both sides of the equation by the coefficient of $x$ , which is $0.8$ :

$10=x$

Therefore, our x-intercept is 10.

8 0

3 years ago

Which expression gives the distance between the points (2,5) and (-4, 8)?

german

Answer:

A. (2+4)² +(5-8)²

Step-by-step explanation:

To find then distance between two points, we will follow the steps below;

write down the formula

D = √(x₂-x₁)²+(y₂-y₁)²

(-4, 8)

x₁=-4

y₁ = 8

(2,5)

x₂=2

y₂=5

we can now proceed to insert the values into the formula

D = √(x₂-x₁)²+(y₂-y₁)²

= √(2+4)²+(5-8)²

=√(6)² + (-3)²

=√36+9

=√45

Therefore the expression which gives the distance between two the two points is (2+4)² +(5-8)²

7 0

3 years ago

Is a relation always a function? Is a function always a relation? Explain.

katen-ka-za [31]

A function is always a relation but relation is not always a function

<u>Solution:</u>

Given that, we have to explain Is a relation always a function and is a function always a relation

Note that both functions and relations are defined as sets of lists.

In fact, every function is a relation. However, not every relation is a function. A relation from a set X to a set Y is called a function if each element of X is related to exactly one element in Y.

That is, given an element x in X, there is only one element in Y that x is related to.

For example, consider the following sets X and Y. Let me give you a relation between them that is not a function;

X = { 1, 2, 3 }

Y = { a , b , c, d }

Relation from X to Y : { (1,a) , (2, b) , (2, c) , (3, d) }

This relation is not a function from X to Y because the element 2 in X is related to two different elements, b and c

Relation from X to Y that is a function: { (1,d) , (2,d) , (3, a) }

This is a function since each element from X is related to only one element in Y. Note that it is okay for two different elements in X to be related to the same element in Y. It's still a function, it's just not a one-to-one function.

So, we can say that function is a type of relation.

Which means whatever a function occurs, it will be a relation from one set to other.

But when a relation occurs it may be a function but need not be always a function.

Hence, a function is always a relation but relation is not always a function.

8 0

3 years ago