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3 years ago

8

I need help on this one

1 answer:

Ivan3 years ago

8 0

All planes without a plane DIH.
Answer:
BGH; CDA; FID; FIH

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17 divided by 3 1/2 Other things ain't working right now so I'm using this.

romanna [79]

Here is your answer friend

6 0

2 years ago

If y varies directly as x, and y=6 as x=-1, find y for the x-value of 5

Tema [17]

Answer:

$y=-30$

Step-by-step explanation:

we are given

y varies directly as x

so, we can write equation as

$y=kx$

where

k is constant of proportionality

We have

at x=-1 , y=6

so, we can use it to find k

$6=k(-1)$

so, we get

$k=-6$

now, we can plug it back

$y=-6x$

now, we can plug x=5

and we get

$y=-6\times 5$

$y=-30$

6 0

3 years ago

Solve for n<br> 3n + 2 - 2n + 5 = 17<br> (Combine like terms then solve.)

elixir [45]

Answer:

n+7=17

n=17-7

n=10

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

David has $65 dollars to spend on a ticket to a concert and a shirt. He already spent $32.25 on his ticket and fee. Find an equa

emmainna [20.7K]

You Only Can Buy Two Shirts To Have Left $3.75

7 0

3 years ago

Evaluate the integral Integral ∫ from (1,2,3 ) to (5, 7,-2 ) y dx + x dy + 4 dz by finding parametric equations for the line seg

n200080 [17]

$\vec F(x,y,z)=y\,\vec\imath+x\,\vec\jmath+3\,\vec k$

is conservative if there is a scalar function $f(x,y,z)$ such that $\nabla f=\vec F$ . This would require

$\dfrac{\partial f}{\partial x}=y$

$\dfrac{\partial f}{\partial y}=x$

$\dfrac{\partial f}{\partial z}=3$

(or perhaps the last partial derivative should be 4 to match up with the integral?)

From these equations we find

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

$\dfrac{\partial f}{\partial y}=x=x+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

$f(x,y,z)=xy+h(z)$

$\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=3z+C$

$f(x,y,z)=xy+3z+C$

so $\vec F$ is indeed conservative, and the gradient theorem (a.k.a. fundamental theorem of calculus for line integrals) applies. The value of the line integral depends only the endpoints:

$\displaystyle\int_{(1,2,3)}^{(5,7,-2)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz=\int_{(1,2,3)}^{(5,7,-2)}\nabla f(x,y,z)\cdot\mathrm d\vec r$

$=f(5,7,-2)-f(1,2,3)=\boxed{18}$

8 0

3 years ago