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ipn [44]
3 years ago
6

Find the directional derivative of the function at the given point in the direction of the vector v. f(x, y, z) = xyz , (3, 3, 9

), v = −1, −2, 2 duf(3, 3, 9)
Mathematics
1 answer:
Anna [14]3 years ago
8 0

The derivative of f(x,y,z) in the direction of a vector \mathbf u is

\nabla_{\mathbf u}f=\nabla f\cdot\dfrac{\mathbf u}{\|\mathbf u\|}

With f(x,y,z)=xyz, we get

\nabla f=(yz,xz,xy)

and \mathbf u=(-1,-2,2),

\|\mathbf u\|=\sqrt{(-1)^2+(-2)^2+2^2}=3

Then

\nabla_{(-1,-2,2)}f(3,3,9)=(27,27,9)\cdot\dfrac{(-1,-2,2)}3=-21

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emmainna [20.7K]
Summarizing the problem, there are three terms that you have to deal with: purchasing cost, down payment and loan. So, you would expect that the answer would contain these quantities. Among them, the unknown is the purchasing cost, therefore, we denote this as x. 

<span>Based on the statement, "The amount of the loan is the purchase cost minus the down payment", we can formulate an equation for this. 

Amount of Loan = x - Down payment

This will be our working equation. Moving on, the down payment was mentioned to be equal to </span>$1500. The lean received is equal to <span>$2600. Substituting these values to the working equation, we can now determine the value of x.

2600 = x - 1500

Solving for x by transposing it to one side,

x = 2600-1500
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3 years ago
How do u write 4 1/4 in simplest form and not in decimals I haven't learned that yet .
alexandr1967 [171]
It already is in simplest form because the numerator and denonimator don't have a common factor to then divide them by.
5 0
2 years ago
Read 2 more answers
Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec
julia-pushkina [17]

Answer:

(2,6,6) \not \in \text{Span}(u,v)

(-9,-2,5)\in \text{Span}(u,v)

Step-by-step explanation:

Let b=(b_1,b_2,b_3) \in \mathbb{R}^3. We have that b\in \text{Span}\{u,v\} if and only if we can find scalars \alpha,\beta \in \mathbb{R} such that \alpha u + \beta v = b. This can be translated to the following equations:

1. -\alpha + 3 \beta = b_1

2.2\alpha+4 \beta = b_2

3. 3 \alpha +2 \beta = b_3

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for \alpha,\beta and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

-\alpha + 3 \beta = 2

2\alpha+4 \beta = 6

whose unique solutions are \alpha =1 = \beta, but note that for this values, the third equation doesn't hold (3+2 = 5 \neq 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

-\alpha + 3 \beta = -9

2\alpha+4 \beta = -2

whose unique solutions are \alpha=3, \beta=-2. Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.

4 0
2 years ago
Solve the problems<br><br>9^2<br><br>4^3<br><br>3^2<br><br>2^3<br><br>5^2
Darina [25.2K]

Answer:

  • 81
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Step-by-step explanation:

hope this helps!

3 0
2 years ago
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myrzilka [38]

Answer:

1/5

Step-by-step explanation:

this is because 4/20 simplifies to 1/5

give brainliest please.

hope this helps :)

7 0
2 years ago
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