Okay here:
Let L be the length and W the width of the rectangle.
1. w=2•L
2. 2L+2W=36
Substitute eq. 1 into eq. 2,
2L+2•(2•L)=36
2L+4L=36
6L=36
L=9
Not done yet, then from eq. 1,
W=2•L
W=2•9
W=18 <------- Your answer. :)
Answer:
A.
Step-by-step explanation:
playing in the garden
Answer:
Step-by-step explanation:
GIVEN: A farmer has 
of fencing to construct a rectangular pen up against the straight side of a barn, using the barn for one side of the pen. The length of the barn is 
.
TO FIND: Determine the dimensions of the rectangle of maximum area that can be enclosed under these conditions.
SOLUTION:
Let the length of rectangle be 
and 
perimeter of rectangular pen 
area of rectangular pen 
putting value of
to maximize 
but the dimensions must be lesser or equal to than that of barn.
therefore maximum length rectangular pen 
width of rectangular pen 
Maximum area of rectangular pen 
Hence maximum area of rectangular pen is 
and dimensions are
Tienes que hallar el mínimo común múltiplo de las 3 cantidades.
18= 2×3²
24= 2³×3
36= 2²×3²
mcm(18,24,36) = 2³×3²=8×9= 72
Eso quier decir que si partieron a la misma hora se encontraran de nuevo en el punto de partida 72 minutos después de la salida.
Las vueltas que habrán realizado será el resultado de dividir 72 entre el tiempo que tardan en dar una vuelta
<span>Mayor: </span> = 4
<span>Mediano: </span> = 3
<span>Pequeño: </span> = 2
Soluciónes:
se vuelven a encontrar a los 72 min de la salida
<span>El mayor dió 4 vueltas, el mediano 3 y el pequeño 2</span>
Vertical shift is -2. Horizontal shift is 1. Check the box that says reflect over x-axis. Horizontal shrink should be 4. Then to check if it looks right go to Desmos.com and in the graphing calculator put in y=-4(x-1)^2-2 and see if it looks like the one you did in your work
