I think 2, 8, 5 <span>those are the factors that can be divided out to simplify the multiplication problem.</span>
Answer:
La cantidad de dinero adeuda al final de los dos procesos es $1450
Step-by-step explanation:
Los parámetros del proceso son;
La cantidad ahorrada para comprar ropa = $2,000
La cantidad pagada por la ropa = $2600
La cantidad prestada a un amigo = $650
La cantidad adeuda por teléfono celular durante 3 meses = $250
Deje que 'C', represente la suma de las transacciones anteriores, tenemos;
∴ C = 2,000 - 2,600 - 650 - 200 = -1,450
La suma de las transacciones anteriores, C = -$1,450
Por lo tanto, al final de los dos procesos, la cantidad de dinero adeuda es C = $1,450
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Answer:
54
Step-by-step explanation:
Answer:
We have sinθ = 12/13
The method here is to figure out the value of θ
Using a calculator sin^(-1)(12/13) =67.38°
67.38° is in quadrant 1 so we must substract 67.38° from 180° wich is π
- 180-67.38= 112.61° ⇒ θ= 112.61°
Now time to calculate cos2θ and cosθ using a calculator
- cosθ = -5/13
- cos2θ = -0.7
The values we got make sense since θ is in quadrant 2 and 2θ in quadrant 3
Answer:

Step-by-step explanation:
We are given the following in the question:
Quantity, q
Selling price in dollars per yard, p

Total revenue earned =

f(20)=13000
This means that 13000 yards of fabric is sold when the selling price is 20 dollars per yard.
f′(20)=−550
This means that increasing the selling price by 1 dollar per yards there is a decrease in fabric sales by 550.
We have to find R'(20)
Differentiating the above expression, we have,

Putting the values, we get,
