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3 years ago

Pls help! I WILL GIVE YOU BRAINLEIST

Mathematics

2 answers:

faltersainse [42]3 years ago

8 0

Answer:

1) = 180 degree reason = being AB is a straight line or supplementary

2) = angle AOB reason = being all the angles are of a straight line AB

3) = x + 2x+34 + 20 = 180 degree. reason being supplementary angle

4) = 42 degree

Nana76 [90]3 years ago

5 0

1) fill 180° in the blank

Reason :

Because AOB angles is half a page so it equals 180°

_________________________________

2)fill 180° in the blank

Reason :

AOE + EOF + FOB = AOB

And we have found that AOB angles is half a page so it equals 180°

_________________________________

3) 2x + 34 _ 180°

_________________________________

4) x = 42°

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Answer:

d = 1.5

Step-by-step explanation:

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Cerrena [4.2K]

By the Pythagorean Theorem:

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4 years ago

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

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4 years ago

Suppose you invest $1,600 at an annual interest rate of 4.6% compounded continuously how much will you have in the account after

DanielleElmas [232]

Total = Principal * e^(rate*years)
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Total = 1,600 * 2.71828182828459^(.184)
Total = 1,600 * 1.2020158231 
Total = 1,923.23 

Source:
http://www.1728.org/rate2.htm

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alekssr [168]

Answer:

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Step-by-step explanation:

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