Answer:
The measure of the arc PJ is ![75\°](https://tex.z-dn.net/?f=75%5C%C2%B0)
Step-by-step explanation:
step 1
Find the measure of angle L
we know that
In a inscribed quadrilateral opposite angles are supplementary
so
![m](https://tex.z-dn.net/?f=m%3CL%2Bm%3CJ%3D180%5C%C2%B0)
we have
![m](https://tex.z-dn.net/?f=m%3CJ%3D110%5C%C2%B0)
substitute
![m](https://tex.z-dn.net/?f=m%3CL%2B110%5C%C2%B0%3D180%5C%C2%B0)
![m](https://tex.z-dn.net/?f=m%3CL%3D70%5C%C2%B0)
step 2
Find the measure of arc KJ
we know that
The inscribed angle measures half that of the arc comprising
so
![m](https://tex.z-dn.net/?f=m%3CP%3D%5Cfrac%7B1%7D%7B2%7D%28arc%5C%20LK%2Barc%5C%20KJ%29)
substitute the values
![95\°=\frac{1}{2}(125\°+arc\ KJ)](https://tex.z-dn.net/?f=95%5C%C2%B0%3D%5Cfrac%7B1%7D%7B2%7D%28125%5C%C2%B0%2Barc%5C%20KJ%29)
![190\°=(125\°+arc\ KJ)](https://tex.z-dn.net/?f=190%5C%C2%B0%3D%28125%5C%C2%B0%2Barc%5C%20KJ%29)
![arc\ KJ=190\°-125\°=65\°](https://tex.z-dn.net/?f=arc%5C%20KJ%3D190%5C%C2%B0-125%5C%C2%B0%3D65%5C%C2%B0)
step 3
Find the measure of arc PJ
we know that
The inscribed angle measures half that of the arc comprising
so
![m](https://tex.z-dn.net/?f=m%3CL%3D%5Cfrac%7B1%7D%7B2%7D%28arc%5C%20PJ%2Barc%5C%20KJ%29)
substitute the values
![70\°=\frac{1}{2}(65\°+arc\ PJ)](https://tex.z-dn.net/?f=70%5C%C2%B0%3D%5Cfrac%7B1%7D%7B2%7D%2865%5C%C2%B0%2Barc%5C%20PJ%29)
![140\°=(65\°+arc\ PJ)](https://tex.z-dn.net/?f=140%5C%C2%B0%3D%2865%5C%C2%B0%2Barc%5C%20PJ%29)
![arc\ PJ=140\°-65\°=75\°](https://tex.z-dn.net/?f=arc%5C%20PJ%3D140%5C%C2%B0-65%5C%C2%B0%3D75%5C%C2%B0)
Answer:
Ty What Do i Do Tho And Btw Ty For The Brainliest If You Give Me Some Have A Nice Rest Of Your Day. From: Destiny To: You ....
Step-by-step explanation:
44 it's above .5 so you round up!
Answer with Step-by-step explanation:
Since 1 mole of sand will contain Avagadro's Number of sand particles (by definition of 1 mole)
Thus we have
1 mole of sand =
sand particles
Thus in number of trillion sand particles we have no of trillion sand particles in 12 mole is
![\frac{6.022\times 10^{23}}{10^{12}}=6.022\times 10^{11}](https://tex.z-dn.net/?f=%5Cfrac%7B6.022%5Ctimes%2010%5E%7B23%7D%7D%7B10%5E%7B12%7D%7D%3D6.022%5Ctimes%2010%5E%7B11%7D)
Now since it is given that mass of 1 trillion sand particles is 1000 tonnes Thus the mass of
trillion sand particles is
![Mass=1000\times 6.022\times 10^{11}=6.022\times 10^{14}tonnes](https://tex.z-dn.net/?f=Mass%3D1000%5Ctimes%206.022%5Ctimes%2010%5E%7B11%7D%3D6.022%5Ctimes%2010%5E%7B14%7Dtonnes)
Part 2)
Since it is given that volume of 1 sand particle is
thus the volume of 1 mole of sand is volume of
sand particles
Thus volume of 1 mole is ![V=1.00\times 10^{-18}km^{3}\times 6.022\times 10^{23}=6.022\times 10^{5}km^{3}](https://tex.z-dn.net/?f=V%3D1.00%5Ctimes%2010%5E%7B-18%7Dkm%5E%7B3%7D%5Ctimes%206.022%5Ctimes%2010%5E%7B23%7D%3D6.022%5Ctimes%2010%5E%7B5%7Dkm%5E%7B3%7D)
Now since the Area of united states is ![A=3.6\times 10^{6}mile^{2}=5.8\times 10^{6}km^{2}](https://tex.z-dn.net/?f=A%3D3.6%5Ctimes%2010%5E%7B6%7Dmile%5E%7B2%7D%3D5.8%5Ctimes%2010%5E%7B6%7Dkm%5E%7B2%7D)
Thus the depth of the sand pile is
![Depth=\frac{Volume}{Area}=\frac{6.022\times 10^5}{5.8\times 10^6}=0.10387km=103.8meters](https://tex.z-dn.net/?f=Depth%3D%5Cfrac%7BVolume%7D%7BArea%7D%3D%5Cfrac%7B6.022%5Ctimes%2010%5E5%7D%7B5.8%5Ctimes%2010%5E6%7D%3D0.10387km%3D103.8meters)
Answer:
<h2>x+4²×3</h2>
<h2>mark me as brainliest</h2>