Angles are undefined terms in plane geometry.
The true statements are:
- <em>4. </em>
<em>.</em> - <em>5. </em>
<em /> - <em>6. </em>
<em> is an obtuse angle</em> - <em>7. Its supplement must be an acute angle</em>
- <em>8. See attachment</em>
<em />
<u>Question 4</u>
The first statement is represented as:
![\angle A \cong \angle B](https://tex.z-dn.net/?f=%5Cangle%20A%20%5Ccong%20%5Cangle%20B)
Supplement of ![\angle B = 22^o](https://tex.z-dn.net/?f=%5Cangle%20B%20%3D%2022%5Eo)
means that A and B are congruent.
The supplementary of A and B would be the same
The measure of A is then calculated as:
![\angle A + 22 = 180](https://tex.z-dn.net/?f=%5Cangle%20A%20%2B%2022%20%3D%20180)
Subtract 22 from both sides
![\angle A = 180 -22](https://tex.z-dn.net/?f=%5Cangle%20A%20%3D%20180%20-22)
![\angle A = 158](https://tex.z-dn.net/?f=%5Cangle%20A%20%3D%20158)
<u>Question 5</u>
If LP and LQ are linear pairs, then:
![LP + LQ = 90](https://tex.z-dn.net/?f=LP%20%2B%20LQ%20%3D%2090)
So:
![\angle Q + LP + LQ = 180](https://tex.z-dn.net/?f=%5Cangle%20Q%20%2B%20LP%20%2B%20LQ%20%3D%20180)
Substitute ![LP + LQ = 90](https://tex.z-dn.net/?f=LP%20%2B%20LQ%20%3D%2090)
![\angle Q + 90 = 180](https://tex.z-dn.net/?f=%5Cangle%20Q%20%2B%2090%20%3D%20180)
Subtract 90 from both sides
![\angle Q =- 90 + 180](https://tex.z-dn.net/?f=%5Cangle%20Q%20%3D-%2090%20%2B%20180)
![\angle Q =90](https://tex.z-dn.net/?f=%5Cangle%20Q%20%3D90)
<u>Question 6</u>
--- S and T are supplementary
--- T and U are complementary
Subtract both equations
![\angle S + \angle T - \angle T - \angle U = 90 - 180](https://tex.z-dn.net/?f=%5Cangle%20S%20%2B%20%5Cangle%20T%20-%20%5Cangle%20T%20-%20%5Cangle%20U%20%3D%2090%20-%20180)
![\angle S - \angle U = -90](https://tex.z-dn.net/?f=%5Cangle%20S%20-%20%5Cangle%20U%20%3D%20-90)
Make U the subject
![\angle U = 90 + \angle S](https://tex.z-dn.net/?f=%5Cangle%20U%20%3D%2090%20%2B%20%5Cangle%20S)
This means that:
![\angle U > 90](https://tex.z-dn.net/?f=%5Cangle%20U%20%3E%2090)
Hence,
is an obtuse angle
<u>Question 7</u>
Obtuse angles are greater than 90 degrees but less than 180.
This means that their supplement must be an acute angle (i.e. less than 90)
<u>Question 8</u>
See attachment
Read more about angles at:
brainly.com/question/13954458
Answer:
words per minute!
Step-by-step explanation:
By the divergence theorem, the flux of
across <em>S</em> is equal to the volume integral of
over the interior of <em>S</em>.
We have
![\vec F(x,y,z) = (x^3+y^3)\,\vec\imath + (y^3+z^3)\,\vec\jmath + (z^3+x^3)\,\vec k \\\\ \implies \mathrm{div}(\vec F) = \dfrac{\partial(x^3+y^3)}{\partial x} + \dfrac{\partial(y^3+z^3)}{\partial y} + \dfrac{\partial(z^3+x^3)}{\partial z} = 3(x^2+y^2+z^2)](https://tex.z-dn.net/?f=%5Cvec%20F%28x%2Cy%2Cz%29%20%3D%20%28x%5E3%2By%5E3%29%5C%2C%5Cvec%5Cimath%20%2B%20%28y%5E3%2Bz%5E3%29%5C%2C%5Cvec%5Cjmath%20%2B%20%28z%5E3%2Bx%5E3%29%5C%2C%5Cvec%20k%20%5C%5C%5C%5C%20%5Cimplies%20%5Cmathrm%7Bdiv%7D%28%5Cvec%20F%29%20%3D%20%5Cdfrac%7B%5Cpartial%28x%5E3%2By%5E3%29%7D%7B%5Cpartial%20x%7D%20%2B%20%5Cdfrac%7B%5Cpartial%28y%5E3%2Bz%5E3%29%7D%7B%5Cpartial%20y%7D%20%2B%20%5Cdfrac%7B%5Cpartial%28z%5E3%2Bx%5E3%29%7D%7B%5Cpartial%20z%7D%20%3D%203%28x%5E2%2By%5E2%2Bz%5E2%29)
so that
![\displaystyle \iint_S \vec F(x,y,z)\cdot\mathrm d\vec s = \iiint_T \mathrm{div}(\vec F)\,\mathrm dV = 3 \iiint\limits_{x^2+y^2+z^2\le3} (x^2+y^2+z^2)\,\mathrm dx\,\mathrm dy\,\mathrm dz](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciint_S%20%5Cvec%20F%28x%2Cy%2Cz%29%5Ccdot%5Cmathrm%20d%5Cvec%20s%20%3D%20%5Ciiint_T%20%5Cmathrm%7Bdiv%7D%28%5Cvec%20F%29%5C%2C%5Cmathrm%20dV%20%3D%203%20%5Ciiint%5Climits_%7Bx%5E2%2By%5E2%2Bz%5E2%5Cle3%7D%20%28x%5E2%2By%5E2%2Bz%5E2%29%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dz)
To compute the volume integral, convert to spherical coordinates with
<em>x</em> = <em>ρ</em> cos(<em>θ</em>) sin(<em>ϕ</em>)
<em>y</em> = <em>ρ</em> sin(<em>θ</em>) sin(<em>ϕ</em>)
<em>z</em> = <em>ρ</em> cos(<em>ϕ</em>)
so that
<em>ρ </em>² = <em>x</em> ² + <em>y</em> ² + <em>z</em> ²
d<em>x</em> d<em>y</em> d<em>z</em> = <em>ρ </em>² sin(<em>ϕ</em>) d<em>ρ</em> d<em>ϕ</em> d<em>θ</em>
The region <em>T</em> is the interior of the sphere <em>S</em>, given by the set
![T = \left\{(\rho,\theta,\phi) \mid 0\le\rho\le3 \text{ and } 0\le\phi\le\pi \text{ and }0\le \theta\le2\pi\right\}](https://tex.z-dn.net/?f=T%20%3D%20%5Cleft%5C%7B%28%5Crho%2C%5Ctheta%2C%5Cphi%29%20%5Cmid%200%5Cle%5Crho%5Cle3%20%5Ctext%7B%20and%20%7D%200%5Cle%5Cphi%5Cle%5Cpi%20%5Ctext%7B%20and%20%7D0%5Cle%20%5Ctheta%5Cle2%5Cpi%5Cright%5C%7D)
So we have
![\displaystyle 3 \int_0^{2\pi} \int_0^\pi \int_0^3 \rho^4 \sin(\phi) \,\mathrm d\rho \,\mathrm d\phi \,\mathrm d\theta \\\\ = 6\pi \left(\int_0^\pi \sin(\phi)\,\mathrm d\phi\right) \left(\int_0^3 \rho^4 \,\mathrm d\rho\right) = \boxed{\frac{2916\pi}5}](https://tex.z-dn.net/?f=%5Cdisplaystyle%203%20%5Cint_0%5E%7B2%5Cpi%7D%20%5Cint_0%5E%5Cpi%20%5Cint_0%5E3%20%5Crho%5E4%20%5Csin%28%5Cphi%29%20%5C%2C%5Cmathrm%20d%5Crho%20%5C%2C%5Cmathrm%20d%5Cphi%20%5C%2C%5Cmathrm%20d%5Ctheta%20%5C%5C%5C%5C%20%3D%206%5Cpi%20%5Cleft%28%5Cint_0%5E%5Cpi%20%5Csin%28%5Cphi%29%5C%2C%5Cmathrm%20d%5Cphi%5Cright%29%20%5Cleft%28%5Cint_0%5E3%20%5Crho%5E4%20%5C%2C%5Cmathrm%20d%5Crho%5Cright%29%20%3D%20%5Cboxed%7B%5Cfrac%7B2916%5Cpi%7D5%7D)
Answer:
C
Step-by-step explanation:
C because they are all even! (Meaning if spun it around, it has an even chance to land on any color)
From bottom to top - corn syrup, glycerol, wafer and corn oil at the top, because the densiest ones sink