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2 years ago

Pleasee help me i’ll give you brainliest

Mathematics

2 answers:

makvit [3.9K]2 years ago

8 0

Choose the first option:

x + y = 180 degrees is the relationship that is always true.

Shalnov [3]2 years ago

4 0

Answer:

X and y are supplementary.

Step-by-step explanation:

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(4x + 17)<br> What is the measure of AC ?<br> Enter your answer in the box.<br> (4x – 3.5)

Natasha2012 [34]

Answer: $AC=41\°$

Step-by-step explanation:

<h3> The complete exercise is attached.</h3>

For this exercise it is important to remember that, by definition, if an angle formed by two intersecting chords has its vertex on a circle, this is called an "Inscribed angle".

The following formula is used to calculate the measure of an Inscribed angle:

$Inscribed\ angle = \frac{1}{2}\ Intercepted\ Arc$

You can identify in the figure that:

$Inscribed\ angle=ABC=(4x -3.5)\\\\Intercepted\ Arc=AC=(4x + 17)$

Then you can substitute values into the formula and solve for "x":

$(4x -3.5)= \frac{1}{2}(4x + 17)\\\\8x-7=4x+17\\\\4x=24\\\\x=6$

Then, AC is:

$AC=(4(6)+ 17)\\\\AC=24+17\\\\AC=41\°$

3 0

3 years ago

What is the produce? -9x(5-2x)

Marta_Voda [28]

Answer:

279x

Step-by-step explanation:

5 0

2 years ago

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All toll free numbers are now working? or all toll free numbers is now working?

mel-nik [20]

<span>All toll free numbers are now working</span>

8 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

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

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3 years ago

Which piecewise function is shown on the graph?

Scilla [17]

The very first one probably

9 0

2 years ago

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