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Paraphin [41]
3 years ago
8

At a sandwich shop, a plain sandwich, which costs $3.00, consists of one type of bread, one type of meat, and one type of cheese

. if the shop has 4 different breads, 8 different meats, and 6 different cheeses, how many possible plain sandwiches can be made?
Mathematics
2 answers:
kramer3 years ago
7 0
192 possible plain sandwiches can be made because you would do 4x6x8 which equals 192.
trasher [3.6K]3 years ago
5 0

Answer:

252

Step-by-step explanation:

just got it right.

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PLEASE HELP!
emmainna [20.7K]

Use the given recursion and starting value of x_0 = 2.4 to find x_1 :

x_1 = \dfrac{x_0 + \frac6{x_0}}2 = \dfrac{2.4 + \frac{6}{2.4}}2 = 2.45

Do the same for x_2 and x_3 :

x_2 = \dfrac{x_1 + \frac6{x_1}}2 = \dfrac{2.45 + \frac6{2.45}}2 \approx 2.44949

x_3 = \dfrac{x_2+\frac6{x_2}}2 \approx \dfrac{2.44949 + \frac6{2.44949}}2 \approx \boxed{2.44949}

(That's not a mistake. This just tells you that the 2nd and 3rd iterates are very close together and have at least the same first 5 digits after the decimal.)

5 0
2 years ago
100 POINTS AND BRAINLIEST!!
Galina-37 [17]

Answer:

A

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
2/7 divided by 7/6 brainlyest on the line
raketka [301]
12/49! hope this helps!
6 0
3 years ago
A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int
Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}

By expanding the previous equation:

c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0

Since there is no accumulation within the tank, expression is simplified to this:

c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}, where c(0) = 0 \frac{pounds}{gallon}.

\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}

The solution of this equation is:

c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})

The salt concentration after 8 minutes is:

c(8) = 0.166 \frac{pounds}{gallon}

The instantaneous amount of salt in the tank is:

m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds

3 0
3 years ago
What is the equivalent to 16 to the 3/4 root
pshichka [43]
<span>√<span><span>​<span>2x−5</span></span>​<span>​​</span></span></span>=<span>7             3/4 
 rate if that help </span>
8 0
3 years ago
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