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3 years ago

Three times a number is 4 more than twice the number

Mathematics

1 answer:

Gelneren [198K]3 years ago

5 0

Answer:

3n = 2n + 4

Step-by-step explanation:

That is that as an equation.

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What is n x 2 squared= 14

zepelin [54]

Answer:

n=3.5

Step-by-step explanation:

2²=4

14÷4=3.5

3.5×4=14

3.5×2²=14

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2 years ago

19 dollar fortnite card

Pachacha [2.7K]

Answer:

coollll

Step-by-step explanation:

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2 years ago

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Show how to use the distributive property to simplify 4(5+x)

nalin [4]

Answer:

20+4x

Step-by-step explanation:

Distributive property is quite easy to learn... its just multiplication in an easier form to visualize once you're used to using it.

To first start this problem you want to use the number 4 in the expression 4(5+x). You then want to multiply 4 by both terms inside of the parentheses. So you would do both 4*5 and 4*x. You would then get 20+4x because 4*5=20 and 4*x is just put together as 4x.

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3 years ago

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The volume of a cube is 8 in^3 What is the length of one side?

stepladder [879]

Answer:

2 in

Step-by-step explanation:

Use the cube volume formula, V = s³, where s is the side length

Plug in 8 as V and solve for s:

V = s³

8 = s³

Take the cube root of both sides:

2 = s

So, the length of one side is 2 inches.

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3 years ago

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For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f

butalik [34]

$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

and that for any two vectors $\vec a$ and $\vec b$ , $\vec a\times\vec b=-\vec b\times\vec a$ , and $\vec a\times\vec a=\vec0$ .

The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

$\nabla\times\vec f=\vec0$

which means $\vec f$ is indeed conservative and we can find $f$ .

Integrate both sides of

$\dfrac{\partial f}{\partial y}=2xze^{2xyz}$

with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

$\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}$

$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

and differentiating with respect to $z$ gives

$\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}$

$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

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

$\implies h(z)=C$

for some constant $C$ . So

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

3 0

3 years ago