Answer:
i believe thats correct
Step-by-step explanation:
They are unspecialised cells or stem cells - these have the ability to develop into different cells with different roles
Answer:
4599
Step-by-step explanation:
= 4500(1 + 2.75/100)^0.8
= 4598.730702
= 4599
I hope it helps.
Answer and Step-by-step explanation:
We are given:
A side length of 7
An angle of 30.
We will use trigonometric functions to solve for a.
We will use the sine trig function.
![sin(Angle) = \frac{Opposite (Side)}{Hypotenuse (Side)}](https://tex.z-dn.net/?f=sin%28Angle%29%20%3D%20%5Cfrac%7BOpposite%20%28Side%29%7D%7BHypotenuse%20%28Side%29%7D)
![sin30 = \frac{7}{a}](https://tex.z-dn.net/?f=sin30%20%3D%20%5Cfrac%7B7%7D%7Ba%7D)
Multiply a by both sides, then divide by sin30 on both sides.
![a = \frac{7}{sin30}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B7%7D%7Bsin30%7D)
Now, use a calculator and do 7 divided by sin of 30.
We get <em><u>a</u></em> to be:
14
<u>a is equal to 14</u>
#teamtrees #WAP (Water And Plant)
Btw you look cute in that picture you took
Check the picture below.
now, bear in mind that, we could simply get the area of the whole circle with that radius, and tease out a quarter, because the segment is just using up a quarter of the circle, because the angle made is 90°, and then subtract the triangle from that sector, and what's leftover is the segment.
![\bf \stackrel{\textit{area of the circle}}{A=\pi r^2}\implies A=\pi (3\sqrt{2})^2\implies A=\pi 3^2\sqrt{2^2}\implies A=18\pi \\\\\\ \textit{now one-quart of that is }\cfrac{18\pi }{4}\implies \cfrac{9\pi }{2}\impliedby \textit{sector's area}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20circle%7D%7D%7BA%3D%5Cpi%20r%5E2%7D%5Cimplies%20A%3D%5Cpi%20%283%5Csqrt%7B2%7D%29%5E2%5Cimplies%20A%3D%5Cpi%203%5E2%5Csqrt%7B2%5E2%7D%5Cimplies%20A%3D18%5Cpi%20%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bnow%20one-quart%20of%20that%20is%20%7D%5Ccfrac%7B18%5Cpi%20%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B9%5Cpi%20%7D%7B2%7D%5Cimpliedby%20%5Ctextit%7Bsector%27s%20area%7D)
![\bf \stackrel{\textit{area of the triangle}}{A=\cfrac{1}{2}bh}\implies A=\cfrac{(3\sqrt{2})(3\sqrt{2})}{2}\implies A=\cfrac{3^2\sqrt{2^2}}{2}\\\\\\A=\cfrac{18}{2}\implies A=9\impliedby \textit{area of that triangle}\\\\ -------------------------------\\\\ \stackrel{sector's area}{\cfrac{9\pi }{2}}~~-~~\stackrel{\textit{area of the triangle}}{9}\implies \cfrac{9\pi -18}{2}\impliedby \textit{segment's area}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20triangle%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7Dbh%7D%5Cimplies%20A%3D%5Ccfrac%7B%283%5Csqrt%7B2%7D%29%283%5Csqrt%7B2%7D%29%7D%7B2%7D%5Cimplies%20A%3D%5Ccfrac%7B3%5E2%5Csqrt%7B2%5E2%7D%7D%7B2%7D%5C%5C%5C%5C%5C%5CA%3D%5Ccfrac%7B18%7D%7B2%7D%5Cimplies%20%0AA%3D9%5Cimpliedby%20%5Ctextit%7Barea%20of%20that%20triangle%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Cstackrel%7Bsector%27s%20area%7D%7B%5Ccfrac%7B9%5Cpi%20%7D%7B2%7D%7D~~-~~%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20triangle%7D%7D%7B9%7D%5Cimplies%20%5Ccfrac%7B9%5Cpi%20-18%7D%7B2%7D%5Cimpliedby%20%5Ctextit%7Bsegment%27s%20area%7D)
or you can always just use the area of a segment, with the radius and angle given.