Answer:
The answer is
or
.
Step-by-step explanation:
Solve the following equation:
![23 \times \frac{1}{5}](https://tex.z-dn.net/?f=23%20%5Ctimes%20%5Cfrac%7B1%7D%7B5%7D)
-Multiply
and
to get
![23 \times \frac{1}{5} = \frac{23}{5} = 4.6](https://tex.z-dn.net/?f=23%20%5Ctimes%20%5Cfrac%7B1%7D%7B5%7D%20%3D%20%5Cfrac%7B23%7D%7B5%7D%20%3D%204.6)
So, now you have found the answer.
9514 1404 393
Answer:
- 13 ft
- (a) 1 second; (b) t = 0, t = 1/2
Step-by-step explanation:
<h3>1. </h3>
Let w represent the length of the wire. Then the height of attachment is (w-1). The Pythagorean theorem tells us a relevant relation is ...
5² +(w -1)² = w²
w² -2w +26 = w² . . . . . . . eliminate parentheses, collect terms
26 = 2w . . . . . . . . . . . . add 2w
13 = w . . . . . . . . . . . . divide by 2
The length of the wire is 13 feet.
__
<h3>2. </h3>
(a) When h = 0, the equation is ...
0 = -16t^2 +8t +8
Dividing by -8 puts this into standard form:
2t^2 -t -1 = 0
Factoring, we get ...
(2t +1)(t -1) = 0
The positive value of t that makes a factor zero is t = 1.
It will take 1 second for the gymnast to reach the ground.
__
(b) When h = 8, the equation is ...
8 = -16t^2 +8t +8
Subtract 8 and divide by 8 to get ...
0 = -2t^2 +t
0 = t(1 -2t) . . . . factor out t
Values of t that make the factors zero are ...
t = 0
t = 1/2
The gymnast will be 8 feet above the ground at the start of the dismount, and 1/2 second later.
Check the picture below.
a)
so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.
the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".
let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.
and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.
![\bf \begin{array}{cllll} \textit{circumference of a circle}\\\\ 2\pi r \end{array}\qquad \qquad \qquad \qquad \begin{array}{cllll} \textit{arc's length}\\\\ s=\cfrac{\theta r\pi }{180} \end{array}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bcllll%7D%0A%5Ctextit%7Bcircumference%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A2%5Cpi%20r%0A%5Cend%7Barray%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Bcllll%7D%0A%5Ctextit%7Barc%27s%20length%7D%5C%5C%5C%5C%0As%3D%5Ccfrac%7B%5Ctheta%20r%5Cpi%20%7D%7B180%7D%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+ \stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2} \\\\\\ 15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Cstackrel%7Bgreen~circle%7D%7Bperimeter%7D%7D%7B2%5Cpi%287.5%29%20%7D~-~%5Cstackrel%7B%5Cstackrel%7Bgreen~circle%7D%7Barc%7D%7D%7B%5Ccfrac%7B%28135%29%287.5%29%5Cpi%20%7D%7B180%7D%7D~%2B%0A%5Cstackrel%7B%5Cstackrel%7Bred~section%7D%7Bperimeter%7D%7D%7B2%5Cpi%289.5%29%20%7D~-~%5Cstackrel%7B%5Cstackrel%7Bred~section%7D%7Barc%7D%7D%7B%5Ccfrac%7B%28135%29%289.5%29%5Cpi%20%7D%7B180%7D%7D%2B%5Cstackrel%7Bendpoints%7D%7B2%2B2%7D%0A%5C%5C%5C%5C%5C%5C%0A15%5Cpi%20-%5Ccfrac%7B45%5Cpi%20%7D%7B8%7D%2B19%5Cpi%20-%5Ccfrac%7B57%5Cpi%20%7D%7B8%7D%2B4%5Cimplies%20%5Ccfrac%7B85%5Cpi%20%7D%7B4%7D%2B4%5Cquad%20%5Capprox%20%5Cquad%2070.7588438888)
b)
we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.
![\bf \begin{array}{cllll} \textit{area of a circle}\\\\ \pi r^2 \end{array}\qquad \qquad \qquad \qquad \begin{array}{cllll} \textit{area of a sector of a circle}\\\\ s=\cfrac{\theta r^2\pi }{360} \end{array}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bcllll%7D%0A%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%5Cpi%20r%5E2%0A%5Cend%7Barray%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Bcllll%7D%0A%5Ctextit%7Barea%20of%20a%20sector%20of%20a%20circle%7D%5C%5C%5C%5C%0As%3D%5Ccfrac%7B%5Ctheta%20r%5E2%5Cpi%20%7D%7B360%7D%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-------------------------------)
Answer: 25
Step-by-step explanation:
Both lines are being cut by the same transversal so to solve for x, make both equations equal to each other
2x+10 = 3x-15
To isolate x, do the inverse operations (for any side chosen)
Do the inverse operations to move -15 and isolate 3x
Add 15 to 10 (10+15=25)
You now have 2x+25=3x
Now, do the same to combine the x values together
25=3x-2x
You now have 25=1x
Divide by 1 on both sides
25=x