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2 years ago

please help!!! ive been trying this question for the pass 20 minutes and cant seem to understand it!!

Mathematics

2 answers:

Lesechka [4]2 years ago

5 0

Answer:

Step-by-step explanation:

Beacuse ‍♀️

mestny [16]2 years ago

4 0

Answer:

This is a rectangle, so if AB is equal to 3 centimeters, then DC is equal to 3 centimeters.

Since Angle X is equal to 45 degrees, then angle CDE, is also equal to 45 degrees, and we have an isosceles triangle, since angle CDE and DEC are equal to each other.

This means, that CE = DC.

And because E is the midpoint of CB, CE = EB.

So basically, we have:

DC = CE = BE

DC = 3

3 = CE = BE

And we see that CE and BE are both equal to 3, which makes CB equal to 6, which makes solving the area easy. (3 x 6), which is 18.

B is correct!

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Hi can someone please help me with this question

olga2289 [7]

Answer: B. 3x + 1/5

tom's pencil is longer than Ellen's pencil:

5x + 2/5 - (2x + 1/5) = 5x - 2x + 2/5 - 1/5 = 3x + 1/5 (cm)

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Find the value of x in each case.

Leto [7]

Answer:

I think its 80 for both. .,.

Hola,

Goodluck! Z:

3 0

2 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

kennys weight is 6/7 times melvins weigh what is the ratio of kennys weight to melvints weight give your answer in fraction form

pychu [463]

Answer:

she ways alot sorry to say

3 0

2 years ago

Find the median of 30 , 40, 35, 50 and 55 degrees angle measures

ELEN [110]

We know that
the median is the middle number when listed in order from least to greatest.

step 1
listed in order from least to greatest
[30,40, 35, 50, 55]--------> [30, 35, 40, 50, 55]

step 2
find the middle number

the middle number is 40

therefore

the answer is
the median is 40°

6 0

3 years ago