Answer:
We have 12 mL of 11% alcohol.
How much water do we add to make it a 6% solution?
We have 1.32 mL alcohol and 10.68 mL water
x is the water to add
(1.32) / (12 + x) = .06
1.32 = .72 + .06x
.6 = .06x
x = 10 mL
Step-by-step explanation:
1/4 hour = 75 gallons
1 hour = 75 x 4 = 300 gallons
--------------------------------------------------------
Answer: The rate is 300 gallons/hour
--------------------------------------------------------
Answer is 3x -2
Kim's age is 3 multiply x=3x
since Sam is 2 years less
Sam's age= 3x-2
Calcula dos números tales que la suma del primero mas el triple del segundo sea 85 y que la diferencia entre el cuádruple del primero y el doble del segundo sea 102
Sea el primer número = T
Sea el segundo número = U
Las ecuaciones según el enunciado son:
1) T + 3U = 85
2) 4T - 2U = 102
Resolvemos por el MÉTODO DE IGUALACIÓN.
-Despejamos T en las dos ecuaciones.
T + 3U = 85 4T - 2U = 102
T = 85 - 3U 4T = 102 + 2U
T = (85 - 3U)/1 T = (102 + 2U)/4
Igualamos las dos ecuaciones entre sí.
(85 - 3U)/1 = (102 + 2U)/4
Multiplicamos las ecuaciones en cruz.
4 (85 - 3u) = 1 (102 + 2U)
340 - 12U = 102 + 2U
- 12U - 2U = 102 - 340
- 14U = - 238
U = - 238/-14
U = 17
El valor de U lo reemplazamos en uno de los despeje de T para hallar el valor de la misma.
T = (85 - 3U)/1
T = (85 - 3 (17)/1
T = (85 - 51)/1
T = 34/1
T = 34
Rpt. Los números son: 34 y 17
COMPROBAMOS LA SOLUCIÓN.
T + 3U = 85
34 + 3 (17) = 85
34 + 51 = 85
85 = 85
4T - 2U = 102
4 (34) - 2 (17) = 102
136 - 34 = 102
102 = 102
LISTO!
Both speeds (quarter horse's and snail's) are already in mph. This means that to be able to determine how many times is the quarter horse faster than a snail, we just have to divide the quarter horse's speed by the speed of the snail.
r = 53.16 / 0.02 = 2658
Therefore, the quarter horse is 2658 times faster than the snail.
