Hi there!
Our interval is from 0 to 3, with 6 intervals. Thus:
3 ÷ 6 = 0.5, which is our width for each rectangle.
Since n = 6 and we are doing a right-riemann sum, the points we will be plugging in are:
0.5, 1, 1.5, 2, 2.5, 3
Evaluate:
(0.5 · f(0.5)) + (0.5 · f(1)) + (0.5 · f(1.5)) + (0.5 · f(2)) + (0.5 · f(2.5)) + (0.5 · f(3)) =
Simplify:
0.5( -2.75 + (-3) + (-.75) + 4 + 11.25 + 21) = 14.875
So, pretend this is your x-axis and y-axis:
I
I
(-2,7) • I
I
I • (2, 5)
I
I
I
I
_________________I____________________
I
I
I
TO GET FROM POINT (-2, 7) TO POINT (2, 5), WE MOVE DOWN 2 AND OVER 4, SO THE SLOPE IS -1/2. IF WE FOLLOW THAT SLOPE AND MOVE DOWN 1 AND OVER 2 FROM THE FIRST POINT OF (-2, 7), WE WILL LAND ON A POINT LOCATED AT (0, 6), WHICH WOULD BE THE "Y-INTERCEPT". WE WERE JUST ABLE TO CALCULATE THE SLOPE OF THE LINE AND THEN USE THE SLOPE TO FIND THE INTERCEPT. SO, THE "SLOPE-INTERCEPT" FORM OF THE EQUATION FOR THIS LINE IS:
y = -1/2x + 6
TO RE-WRITE THIS IN STANDARD FORM, WE JUST WANT TO MOVE THE X VARIABLE OVER TO THE LEFT WITH THE Y VARIABLE, SO:
y = -1/2x + 6
+1/2x + 1/2x
1/2x + y = 6 .... and that is your answer!
3.33 repeating. if you look at it it's 2.70 what I would do is take away the zero for a minute and see how many times 9 goes into 27
Answer:
El área del cuadrado es de 3 centímetros cuadrados.
Step-by-step explanation:
Dado que se desea pintar un cuadrado inscrito en una circunferencia de radio R=3cm como se muestra en la figura, para calcular el área del cuadrado se debe realizar el siguiente cálculo:
Radio = diámetro / 2
3 = X/2
X = 6
Hipotenusa del triángulo interno: 6 cm
Aplicando teorema pitagórico: lado al cuadrado mas lado al cuadrado es igual a hipotenusa al cuadrado.
L^2 + L^2 = 6
2L^2 = 6
L^2 = 6/2
L = √ 3
L = 1.732
Área de un cuadrado = L x L = L^2 = 3
Por lo tanto, el área del cuadrado es de 3 centímetros cuadrados.
Answer:
okay , attach a file .
Step-by-step explanation:
