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Wittaler [7]
3 years ago
9

Solve z=2/11c for c

Mathematics
1 answer:
Sonja [21]3 years ago
5 0
Z= 2 / 11 c

z • 11/2 = c
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Just need help with #2 <br> Please help!<br> Thank you <br> (15 points)
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So you write into the table that x = 1 then y will equal 1*1.5 which equal to 1.5. So under x = 1 in y table you write y = 1.5, and so on

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Paul's bathtub is clogged he has to empty 30 liters of water by hand. Paul has a 3-liter, a 4-liter, and a 5-liter bucket. if Pa
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3 years ago
Y are selling lollipops for $0.50 and candy bars for $1.50.
frutty [35]

Answer:

b=8

Step-by-step explanation:

You multiply $0.50 by 6 (to get the amount of money she spent on lollipops)=3 then you take the 3 and subtract it from $15 which gives you 12, finally you get 1.50-12 and do that 8 times which means she bought 8 candy bars

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2 years ago
Given the following function ,find h(-4) <br><br>h(x)=6x^2+5​
Wittaler [7]

Answer:

h(-4) =101

Step-by-step explanation:

H(x) = 6x ^2 +5

We have  to find when h=4, putting it into function

h(-4) = 6(-4) ^2 +5

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3 0
3 years ago
If a cup of coffee has temperature 97°C in a room where the ambient air temperature is 25°C, then, according to Newton's Law o
FinnZ [79.3K]

Answer:

The  average temperature  is T_{a} = 81.95^oC

Step-by-step explanation:

From the question we are told that

    The temperature of the coffee after time t is   T(t) =  25 + 72 e^{[-\frac{t}{45} ]}

Now the average temperature during the first 22 minutes i.e fro 0 \to  22minutes is mathematically evaluated as

              T_{a} =  \frac{1}{22-0}  \int\limits^{22}_{0} {25 +72 e^{[-\frac{t}{45} ]}} \, dx

               T_{a} = \frac{1}{22} [25 t  +  72 [\frac{e^{[-\frac{t}{45} ]}}{-\frac{1}{45} } ] ] \left| 22} \atop {0}} \right.

             T_{a} = \frac{1}{22} [25 t  - 3240e^{[-\frac{t}{45} ]} ] \left | 45} \atop {{0}} \right.

              T_{a} = \frac{1}{22} [25 (22)  - 3240e^{[-\frac{22}{45} ]}   - (- 3240e^{0} )]

            T_{a} = \frac{1}{22} [550  - 1987.12  +  3240]

          T_{a} = 81.95^oC

       

8 0
2 years ago
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