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wlad13 [49]
3 years ago
12

A medium sized apple has 70 calories. This is 10 calories less than 1\4 of the calories in an old westie chocolate bar.How many

calories are in the chocolate bar?What is the best model for yourr first step
Mathematics
2 answers:
n200080 [17]3 years ago
7 0

Answer:

The answer to your question is: 320

Step-by-step explanation:

Data

1/4 of the calories in a chocolate - 10 = calories in an apple

Equation

    Calories in a chocolate bar = 4(calories in apple + ten)

                                                                    = 4(70 + 10)

                                                                    = 4(80)

                                                                   = 320

                     

Paha777 [63]3 years ago
6 0

Answer:

320 calories.

Step-by-step explanation:

Let the number of calories in the chocolate bar be x, then:

1/4 x - 10 = 70

1/4x = 80

Multiply both sides by 4:

x = 320 calories.

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Kevin and Randy Muise have a jar containing 36 ​coins, all of which are either quarters or nickels. The total value of the coins
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3 years ago
) Picture frames are produced such that the four sides of a picture frame consist of two pieces from a population whose mean len
Natali [406]

Answer:

The mean perimeter of frame = 38 cm

Step-by-step explanation:

By perimeter of a rectangle, the formula is given as:

Perimeter of rectangle = 2(A + B)

Going by the illustration in the question, the perimeter is

==> 2(32 + 44) = 152 cm.

However, we a given the mean perimeter of the frame which are:

32 and 44. We can either choose to say; each sides is (32 by 32) and (44 by 44). Or we consider it in mean/average term as given: (32 by 44).

Case 1: To obtain the mean perimeter of the frame using the first scenario, we say:

Mean perimeter = (32+32+44+44)/4 = 38 cm.

Case 2: We can simply say:

Mean perimeter = (32 + 44)/2 = 38 cm.

Graphically: Assuming that the frame is given below

                     32 cm

          ---------------------------------

         |                                      |

         |                                      |

44 cm|                                      | 44 cm

         |                                      |

         |                                      |

         |                                      |

         ----------------------------------

                       32 cm

The mean of upper frame/bars = 32 cm

The mean of side frame/bars  =  44 cm

Hence, the mean of the perimeter = (32 + 44)/2 = 38 cm

OR,

we take the sum of all the sides (32 + 44 + 32 + 44) and divide by total number of sides (4). That is;

The mean of the perimeter = (32 + 44 + 32 + 44)/4 = 38 cm

3 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
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