We develop an equation for the given situation by first writing the general equation for lines,
y = mx + b
Substituting to this given the values given above,
(1990) 430 = b
(2000) 400 = m(10) + 430
The value of m from the equation in 2000 is -3. Thus, the equation of that relates the variables is,
y = -3x + 430
Answer:
x = number of nickels = 127
y = number of dimes = 156
z = number of quarters = 78
Step-by-step explanation:
Let
x = number of nickels
y = number of dimes
z = number of quarters
Total worth of the coins = $41.45
Total number of coins = 361
x + y + z = 361 (1)
dime = $0.1,
nickel = $0.05
quarter = $0.25
0.05x + 0.1y + 0.25z = 41.45 (2)
twice as many dimes as quarters.
y = 2z
Substitute y = 2z into (1) and (2)
x + 2z + z = 361
0.05x + 0.1(2z) + 0.25z = 41.45
x + 3z = 361
0.05x + 0.2z + 0.25z = 41.45
x + 3z = 361 (3)
0.05x + 0.45z = 41.45 (4)
Multiply (4) by 20
x + 3z = 361 (3)
x + 9z = 829 (5)
Subtract (3) from (5)
9z - 3z = 829 - 361
6z = 468
Divide both sides by 6
z = 468 / 6
= 78
z= 78
Recall,
y = 2z
= 2(78)
= 156
y = 156
Substitute the value of y and z into
x + y + z = 361
x + 156 + 78 = 361
x + 234 = 361
x = 361 - 234
= 127
x= 127
x = number of nickels = 127
y = number of dimes = 156
z = number of quarters = 78
It all depends on how big the boxes are but i would say about 5 or 6 boxes if they are big boxes
Answer:
2224.32 cm³
Step-by-step explanation:
La fórmula para el volumen de una pirámide hexagonal = 1/3 (1/2 × P × a) × h
Donde a = Apotema = 9,93 cm
P = Perímetro =
Pirámide hexagonal = 6 × Bordes de la base = 6 × 8 cm = 48 cm
h = Altura = 28 cm
Por eso,
Volumen de la pirámide hexagonal = 1/3 (1/2 (48 × 9.93) × 28 cm
= 1/3 × 238.32 × 28 cm
= 2224.32 cm³
Volumen de la pirámide hexagonal = 2224.32 cm³
