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borishaifa [10]

3 years ago

15

Find the distance from A to C

1 answer:

Vedmedyk [2.9K]3 years ago

5 0

Answer: 11 because the distance between A-B is 8 so because it says it on top, then add 3 to it from B to C

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Identify whether the following equation has a unique solution, no solution, or infinitely many solutions.

lidiya [134]

Answer:

No solution

Step-by-step explanation:

3+6=9

3-4=-1

9 doesn't equal -1 therefore no solution.

3 0

3 years ago

EXAMPLE 5 If F(x, y, z) = 4y2i + (8xy + 4e4z)j + 16ye4zk, find a function f such that ∇f = F. SOLUTION If there is such a functi

Valentin [98]

If there is such a scalar function <em>f</em>, then

$\dfrac{\partial f}{\partial x}=4y^2$

$\dfrac{\partial f}{\partial y}=8xy+4e^{4z}$

$\dfrac{\partial f}{\partial z}=16ye^{4z}$

Integrate both sides of the first equation with respect to <em>x</em> :

$f(x,y,z)=4xy^2+g(y,z)$

Differentiate both sides with respect to <em>y</em> :

$\dfrac{\partial f}{\partial y}=8xy+4e^{4z}=8xy+\dfrac{\partial g}{\partial y}$

$\implies\dfrac{\partial g}{\partial y}=4e^{4z}$

Integrate both sides with respect to <em>y</em> :

$g(y,z)=4ye^{4z}+h(z)$

Plug this into the equation above with <em>f</em> , then differentiate both sides with respect to <em>z</em> :

$f(x,y,z)=4xy^2+4ye^{4z}+h(z)$

$\dfrac{\partial f}{\partial z}=16ye^{4z}=16ye^{4z}+\dfrac{\mathrm dh}{\mathrm dz}$

$\implies\dfrac{\mathrm dh}{\mathrm dz}=0$

Integrate both sides with respect to <em>z</em> :

$h(z)=C$

So we end up with

$\boxed{f(x,y,z)=4xy^2+4ye^{4z}+C}$

7 0

3 years ago

What is<br> 15/100<br> as a percentage?

Lunna [17]

Answer:

15%

Step-by-step explanation:

hope i helped!! please follow ❤❤

7 0

2 years ago

Read 2 more answers

Carmen wins $35 was in the contest. Tubes of her favorite brand of paint cost $4.80 each. Paint brushes cost $6.70 each. How man

Zanzabum

6.70 for paint brushes. for 2 is 6.70×2=$13.40

35-13.40=$21.60 left after 2 paint brushes

4.80×4=$19.20 for 4 tubes of paint

21.60-19.20=$2.40 left after 4 tubes of paint and 2 paint brushes

6 0

3 years ago

This is confusing<br> (2x+y)^2−(x−2y)^2

Iteru [2.4K]

Answer:

${(2x + y)}^{2} - {(x - 2y)}^{2}$

$< {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} >$

$< {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} >$

${(2x)}^{2} + 2(2x)(y) + {(y)}^{2} - {(x)}^{2} - 2(x)(2y) + {(2y)}^{2}$

${4x}^{2} + 4xy + {y}^{2} - {x}^{2} - 4xy + {4y}^{2}$

${4x}^{2} - {x}^{2} + {4y}^{2} + {y}^{2}$

$= {3x}^{2} + {5y}^{2}$

5 0

3 years ago

Read 2 more answers