Queremos encontrar las ecuciones de las circunferencias que definen el reloj dado.
Las ecuaciones son:
x^2 + y^2 = 16cm^2
x^2 + y^2 = 196cm^2
Lo primero que debemos hacer para escribir una ecuacion, es definir nuestro eje de coordenadas.
Pondremos el punto (0, 0) en el centro de la cartulina.
Debemos recordar que la ecuación de un circulo de radio R centrado en el punto (a, b) se escribe como:
(x - a)^2 + (y - b)^2 = R^2
En este caso tendremos dos ecuaciones de circulos, ambas centradas en el centro de la cartulina, que es el punto (0, 0).
Para el primer caso tendremos un radio R = 4cm, que es el circulo interno que contiene las horas 12, 3, 6, y 9.
Otro de radio R' = 14cm, que contiene el resto de horas.
Entonces las dos ecuaciones de los circulos seran:
x^2 + y^2 = (4cm)^2 = 16cm^2
x^2 + y^2 = (14cm)^2 = 196cm^2
Si quieres aprender más. puedes leer.
brainly.com/question/24021353
Answer:
Step-by-step explanation:
![\\ \int\limits^{a}_{0} \int\limits^{x}_{0} \int\limits^{x+y}_{0} {e^{x+y+z}} \, dzdydx \\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [\int\limits^{x+y}_{0} {e^{x+y}e^z} \, dz]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}\int\limits^{x+y}_{0} {e^z} \, dz]dydx\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}e^z\Big|_0^{x+y}]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}e^{x+y}-e^{x+y}]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} e^{2x+2y}-e^{x+y}dydx \\\\\\](https://tex.z-dn.net/?f=%5C%5C%20%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20%5Cint%5Climits%5E%7Bx%2By%7D_%7B0%7D%20%7Be%5E%7Bx%2By%2Bz%7D%7D%20%5C%2C%20dzdydx%20%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20%5B%5Cint%5Climits%5E%7Bx%2By%7D_%7B0%7D%20%7Be%5E%7Bx%2By%7De%5Ez%7D%20%5C%2C%20dz%5Ddydx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20%5Be%5E%7Bx%2By%7D%5Cint%5Climits%5E%7Bx%2By%7D_%7B0%7D%20%7Be%5Ez%7D%20%5C%2C%20dz%5Ddydx%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20%5Be%5E%7Bx%2By%7De%5Ez%5CBig%7C_0%5E%7Bx%2By%7D%5Ddydx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20%5Be%5E%7Bx%2By%7De%5E%7Bx%2By%7D-e%5E%7Bx%2By%7D%5Ddydx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20e%5E%7B2x%2B2y%7D-e%5E%7Bx%2By%7Ddydx%20%5C%5C%5C%5C%5C%5C)
![\\=\int\limits^{a}_{0} [\int\limits^{x}_{0} e^{2x}e^{2y}-e^{x+y}dy]dx \\\\\\=\int\limits^{a}_{0} [\int\limits^{x}_{0} e^{2x}e^{2y}dy- \int\limits^{x}_{0}e^{x}e^{y}dy]dx \\\\\\u=2y\\du=2dy\\dy=\frac{1}{2}du\\\\\\=\int\limits^{a}_{0} [\frac{e^{2x}}{2}\int e^{u}du- e^x\int\limits^{x}_{0}e^{y}dy]dx \\\\\\=\int\limits^{a}_{0} [\frac{e^{2x}}{2}\cdot e^{2y}\Big|_0^x- e^xe^{y}\Big|_0^x]dx \\\\\\=\int\limits^{a}_{0} [\frac{e^{2x+2y}}{2} - e^{x+y}\Big|_0^x]dx \\\\](https://tex.z-dn.net/?f=%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5B%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20e%5E%7B2x%7De%5E%7B2y%7D-e%5E%7Bx%2By%7Ddy%5Ddx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5B%5Cint%5Climits%5E%7Bx%7D_%7B0%7D%20e%5E%7B2x%7De%5E%7B2y%7Ddy-%20%5Cint%5Climits%5E%7Bx%7D_%7B0%7De%5E%7Bx%7De%5E%7By%7Ddy%5Ddx%20%5C%5C%5C%5C%5C%5Cu%3D2y%5C%5Cdu%3D2dy%5C%5Cdy%3D%5Cfrac%7B1%7D%7B2%7Ddu%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5B%5Cfrac%7Be%5E%7B2x%7D%7D%7B2%7D%5Cint%20e%5E%7Bu%7Ddu-%20e%5Ex%5Cint%5Climits%5E%7Bx%7D_%7B0%7De%5E%7By%7Ddy%5Ddx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5B%5Cfrac%7Be%5E%7B2x%7D%7D%7B2%7D%5Ccdot%20e%5E%7B2y%7D%5CBig%7C_0%5Ex-%20e%5Exe%5E%7By%7D%5CBig%7C_0%5Ex%5Ddx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5B%5Cfrac%7Be%5E%7B2x%2B2y%7D%7D%7B2%7D%20-%20e%5E%7Bx%2By%7D%5CBig%7C_0%5Ex%5Ddx%20%5C%5C%5C%5C)
![\\=\int\limits^{a}_{0} [\frac{e^{4x}}{2} - e^{2x}-\frac{e^{2x}}{2} + e^{x}]dx \\\\\\=\int\limits^{a}_{0} \frac{e^{4x}}{2} -\frac{3e^{2x}}{2} + e^{x}dx \\\\\\=\int\limits^{a}_{0} \frac{e^{4x}}{2}dx -\int\limits^{a}_{0}\frac{3e^{2x}}{2}dx + \int\limits^{a}_{0}e^{x}dx \\\\\\u_1=4x\\du_1=4dx\\dx=\frac{1}{4}du_1\\\\\u_2=2x\\du_2=2dx\\dx=\frac{1}{2}du_2\\\\\\=\frac{1}{8}\int e^{u_1}du_1 -\frac{3}{4}\int e^{u_2}du_2 + \int\limits^{a}_{0}e^{x}dx \\\\\\](https://tex.z-dn.net/?f=%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5B%5Cfrac%7Be%5E%7B4x%7D%7D%7B2%7D%20-%20e%5E%7B2x%7D-%5Cfrac%7Be%5E%7B2x%7D%7D%7B2%7D%20%2B%20e%5E%7Bx%7D%5Ddx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cfrac%7Be%5E%7B4x%7D%7D%7B2%7D%20-%5Cfrac%7B3e%5E%7B2x%7D%7D%7B2%7D%20%2B%20e%5E%7Bx%7Ddx%20%5C%5C%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%20%5Cfrac%7Be%5E%7B4x%7D%7D%7B2%7Ddx%20-%5Cint%5Climits%5E%7Ba%7D_%7B0%7D%5Cfrac%7B3e%5E%7B2x%7D%7D%7B2%7Ddx%20%2B%20%5Cint%5Climits%5E%7Ba%7D_%7B0%7De%5E%7Bx%7Ddx%20%5C%5C%5C%5C%5C%5Cu_1%3D4x%5C%5Cdu_1%3D4dx%5C%5Cdx%3D%5Cfrac%7B1%7D%7B4%7Ddu_1%5C%5C%5C%5C%5Cu_2%3D2x%5C%5Cdu_2%3D2dx%5C%5Cdx%3D%5Cfrac%7B1%7D%7B2%7Ddu_2%5C%5C%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B8%7D%5Cint%20e%5E%7Bu_1%7Ddu_1%20-%5Cfrac%7B3%7D%7B4%7D%5Cint%20e%5E%7Bu_2%7Ddu_2%20%2B%20%5Cint%5Climits%5E%7Ba%7D_%7B0%7De%5E%7Bx%7Ddx%20%5C%5C%5C%5C%5C%5C)
Sorry if that took a while to finish. I am in AP Calculus BC and that was my first time evaluating a triple integral. You will see some integrals and evaluation signs with blank upper and lower boundaries. I just had my equation in terms of u and didn't want to get any variables confused. Hope this helps you. If you have any questions let me know. Have a nice night.
The answer would be D, I think
Answer:
The slope is $0.35/min and it gives the cost per minute of the phone used.
Step-by-step explanation:
We can model this situation with a linear equation of the form
where 
is monthly cost, 
is the number of minutes, 
is the flat monthly fee, and 
is the slope of the equation, or in our case, the amount of money charged per minute.
The slope is
,
in other words, the phone company charges $0.5 per minute.
With the slope in hand, the linear equation becomes
,
and we can find the monthly fee from that fact that for 300 minutes the cost is $131:
.
Therefore,
where the slope if the equation give the cost per minute of the phone used.
Answer:
Num. 2: A = 25, Num. 3 = 63, Num. 4 = 80
Step-by-step explanation:
I'll help with as many as I can, so a few of them I won't be able too answer. You should re-post those later. Good luck, hope this helps.
For number 2:
First, turn the triangle into a rectangle by cutting it in half and attaching it to the other half so it forms a rectangle.
Since you cut the triangle in half, you also have to split 5 in half, so the bottom length is now 2.5.
Now, simply multiply 2.5 by 10, and you have the area for number 2.
The area for number 2 is 25.
For number 3:
This one is much easier. Just think of the shape as a normal square that has been tipped to the side. That means that you would solve for area the same way: 9 x 7, which means that the area for number 3 is 63.
For number 4:
This one is also easy. Simply cut, and paste. Now the right side is 10, and the top side is 8. Multiply, and the area will be 80 m.
Please re-post 1 and 5, as I am not able to solve them. Sorry this answer took so long!
