Answer:
la distancia del punto de observación A al incendio = 5.5 millas
la distancia del punto de observación B al incendio = 3.8 millas
Step-by-step explanation:
La expresión esquemática de la pregunta se puede ver en la imagen adjunta a continuación :
Del siguiente diagrama;
Usando la sine regla:
a sin 79 = 6 sin 63
a = 5.446 millas
la distancia del punto de observación A al incendio = 5.5 millas (a la décima más cercana)
b sin 79 = 6 sin 38
b = 3.7635 millas
la distancia del punto de observación B al incendio = 3.8 millas (a la décima más cercana)
Answer:
1. Western erodes 2 ft/yr; Dunes builds up at 5 ft/yr
2. Sometime in 2006
3. Solve simultaneous equations
Step-by-step explanation:
1. Erosion patterns
(a) Western Beach
In 15 yr, Western Beach erodes from 100 ft to 70 ft.
The rate of erosion is 30 ft/15 yr = 2 ft/yr.
(b) Dunes Beach
In 15 yr, Dunes Beach builds up from 20 ft to 95 ft.
The rate of buildup is 75 ft/15 yr = 5 ft/yr.
2. Beaches with equal width
From the table, it appears that the beaches will have the same width sometime in year 11 (2006).
3. Best approximation
The graph below also shows that it happens part way through year 11 (2006).
We could get an even better solution by calculating the equations of the two lines and solving the simultaneous equations.
Answer:
Step-by-step explanation:
24
Answer:
The maximum number of pounds of potato salad that Charlie can buy is 0.375
Step-by-step explanation:
see the attached figure to better understand the problem
Let
a ----> the cost of one tuna sandwich
b ----> the cost of a bottle of apple juice
c ----> the cost per pound of potato salad
x ----> pounds of potato salad
we have
we know that
He wants to buy a tuna sandwich, a bottle of apple juice, and x pounds of potato salad and can spend up to $8
The inequality that represent this situation is
substitute the given values
Solve for x
Combine like terms
Subtract 6.50 both sides
Divide by 4 both sides
therefore
The maximum number of pounds of potato salad that Charlie can buy is 0.375
They must sell 250 cameras
and 50 more cameras would be a $400 profit
