All categories

3 years ago

Help-

Mathematics

1 answer:

soldi70 [24.7K]3 years ago

7 0

Answer:

root 49= 7^2

root 100= 10^2

root 144= 12^2

root 324= 18^2

root 441= 21^2

Step-by-step explanation:

You might be interested in

A bag contains 10 blue objects and 10 red objects. There are 5 blue circles, 5 red circles, 5 blue squares and 5 red squares.

Amiraneli [1.4K]

The answer to your question I be alive would be A)

3 0

3 years ago

Read 2 more answers

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

4 years ago

PLEASE HELP QUICK I NEED IT TO PASS

ivanzaharov [21]

816.81
Hope this helps

4 0

3 years ago

3, 6, 12, 24 Find the definition

SashulF [63]

Answer:

Multiply by 2 because if 6÷3=2, same for 24÷12=2

6 0

3 years ago

Read 2 more answers

NO LINKS!!! 14. Consider the table

iVinArrow [24]

Answer:

(a) rate of change = 8

(b) Function: y = 8x

Range: 0 ≤ y ≤ 16

Explanation:

Looks like a inverse variation sequence.

Inverse Variation formula: y = k(x)

Take two points: (1, 8), (2, 16)

Find the value of k which is constant also considered as rate of change.

Insert values:

8 = k(1)

k = 8

===========

16 = k(2)

k = 8

Equation: y = 8x

Domain lies in the x axis, Range lies in the y axis.

Here domain: 0 ≤ x ≤ 2

Here range: 0 ≤ y ≤ 16

5 0

2 years ago

Read 2 more answers