A=45
A=kx/y
where k is constant
substituting the values for k
k= 20×9/6=30
when x=9, y=6
A= 30×9/6=45
Answer:
- Domain:
- Range:
Step-by-step explanation:
The domain is the set of x values and the range is the set of y-values.
Answer:
4
Step-by-step explanation:
This situation has two unknowns - the total number of half dollars and the total number of quarters. Because we have two unknowns, we will write a system of equations with two equations using the two unknowns.
- h+q=31 is an equation representing the total number of coins
- 0.50h+0.25q=11 is an equation representing the total value in money based on the number of coin. 0.50 and 0.25 come from the value of a half dollar and quarter individually.
We write the first equation in terms of q by subtracting it across the equal sign to get h=31-q. We now substitute this for h in the second equation.
0.50(31-q)+0.25q=11
15.5-0.50q+0.25q=11
15.5-0.25q=11
After simplifying, we subtract 15.5 across and divide by the coefficient of q.
-0.25q=-4.5
q=4
We now know of the 31 coins that 4 are quarters.
Answer:
x+41=-11 I guess
Step-by-step explanation:
Answer:
En el curso anterior había 430 alumnos.
Step-by-step explanation:
El curso tiene 473 alumnos. Nos dicen que respecto al curso anterior se ha producido un aumento de inscripciones del 10 %. Entonces, siendo x la cantidad de alumnos que había en el curso anterior, se puede plantear la ecuación:
x + 0.1*x= 473
Resolviendo se obtiene:
1.1*x=473
x= 473 ÷1.1
x= 430
<u><em>En el curso anterior había 430 alumnos.</em></u>
