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2 years ago

The formula for the resistance, R, of a conductor with voltage, V, and current, I, is R = V/I.

Mathematics

2 answers:

Karo-lina-s [1.5K]2 years ago

5 0

Answer:

What can I do for you?

Step-by-step explanation:

Komok [63]2 years ago

3 0

I’m pretty sure the answer is D because you need separate the variable V by itself do you divide I from V and then the division changes to IR. If I’m wrong on that, then it’s A

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Problems 1-2, which symbol should be used to make the values true; > or

dimaraw [331]

Answer:

1. <

2. <

Step-by-step explanation:

Have a good day :)

5 0

2 years ago

A soundproofing tile is made of 8 identical solid right pyramids with square bases. The length of the whole tile shown is x inch

mylen [45]

Answer:

Option B ⇒ Area of base of each pyramid :

$(\frac{x}{4})^2 in^2$

Option A ⇒ Volume of pyramid :

$\frac{1}{3} x^{2} \cdot y$ cm3

Step-by-step explanation:

One side of tile consists of 4 pyramids and since length of whole tile is x,

So, the base length of each pyramid = $\frac{x}{4}$

⇒ Area of base of each pyramid = side²

⇒ Area =

$(\frac{x}{4})^2 in^2$

Volume of square based pyramid :

$=(Area of base )\cdot \frac{h}{3}\\=\frac{x^2}{3} cm^3$

For 1. The correct Option is B

For 2. The correct option is A

7 0

2 years ago

Read 2 more answers

How do you simplify this expression

ehidna [41]

The picture is blurry

3 0

3 years ago

F(x, y, z) = yzexzi + exzj + xyexzk,

Ghella [55]

$\nabla f(x,y,z)=\mathbf F(x,y,z)\iff\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j+\dfrac{\partial f}{\partial z}\,\mathbf k=yze^{xz}\,\mathbf i+e^{xz}\,\mathbf j+xye^{xz}\,\mathbf k$

$\dfrac{\partial f}{\partial x}=yze^{xz}$
$\implies f(x,y,z)=\dfrac{yz}ze^{xz}+g(y,z)=ye^{xz}+g(y,z)$

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

$\dfrac{\partial f}{\partial z}=xye^{xz}=xye^{xz}+\dfrac{\mathrm dh}{\mathrm dz}$
$\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C$

$f(x,y,z)=ye^{xz}+C$

$\mathbf r(t)=(t^2+5)\,\mathbf i+(t^2-1)\,\mathbf j+(t^2-5)\,\mathbf k$
$\implies\mathbf r(0)=5\,\mathbf i-\mathbf j-5\,\mathbf k$
$\implies\mathbf r(5)=30\,\mathbf i+24\,\mathbf j+20\,\mathbf k$

By the gradient theorem, we have for any path $\mathcal C$ originating at the point (5, -1, -5) and terminating at the point (30, 24, 20),

$\displaystyle\int_{\mathcal C}\mathbf F\cdot\mathrm d\mathbf r=f(\mathbf r(5))-f(\mathbf r(0))=f(30,24,20)-f(5,-1,-5)$
$\implies\displaystyle\int_{\mathcal C}\mathbf F\cdot\mathrm d\mathbf r=24e^{600}+e^{-25}$

4 0

3 years ago

3x-4y>12

Anvisha [2.4K]

For this case we have the following inequality:

$3x-4y> 12$