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Vitek1552 [10]
3 years ago
10

Roups

Mathematics
1 answer:
Burka [1]3 years ago
7 0

###### ### ### ##### #### ###### #### #### :)

<em>NOT VALID</em>

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Cookies are on sale! Today each cookie costs $0.75, less than the normal price. Right now if you buy 7 of them it will only cost
Sophie [7]
Equation : c = (2.80 / 7) + 0.75

Normal price of each cookie:
c = (2.80 / 7) + 0.75
c = 0.4 + 0.75
c = $1.15
3 0
4 years ago
Ella has 0.5 lbs of sugar. How much water should she add to make the following concentrations? Tell Ella how much syrup she will
Oksanka [162]

Answer:

Ella should add 32\dfrac{5}{6} lbs of water.

Total weight of syrup 33\dfrac{1}{3} lbs

Step-by-step explanation:

Weight:

Weight of sugar = 0.5 lbs.

Weight of water added = x lbs

Total weight = 0.5 + x lbs

Percentage:

0.5 + x  --  100%

0.5  --  1.5%

Write a proportion:

\dfrac{0.5+x}{0.5}=\dfrac{100}{1.5}

Cross multiply:

1.5(0.5+x)=0.5\cdot 100\\ \\0.75+1.5x=50\\ \\1.5x=50-0.75\\ \\1.5x=49.25\\ \\x=\dfrac{49.25}{1.5}=\dfrac{4,925}{150}=\dfrac{197}{6}=32 \dfrac{5}{6}\ lbs

Ella should add 32\dfrac{5}{6} lbs of water.

Total weight of syrup

32\dfrac{5}{6}+\dfrac{1}{2}=32\dfrac{5}{6}+\dfrac{3}{6}=33\dfrac{1}{3}\ lbs

3 0
3 years ago
Comparing freezers
zvonat [6]

Answer:

Freezer B has 8 more cubic feet than Freezer A.

Step-by-step explanation:

Volume of Freezer A = 8 * 4 * 2 = 64 ft^3.

Volume of Freezer B = 6 * 4 * 3 = 72 ft^3.

72 - 64 = 8 ft^3.

8 0
3 years ago
Urgent help!<br><br> refer to photo
vladimir1956 [14]
The answer is a, a=15
7 0
3 years ago
A cold drink is poured out at 52°F. After 2 minutes of sitting in a 72°F room, its temperature has risen to 55°F. Find an equati
I am Lyosha [343]

Answer:

The model for the temperature of the drink can be written as

T=72-20e^{-0.08t}

Step-by-step explanation:

For a cold drink in a hotter room, we can say that the rate of change of temperature of the drink is proportional to the difference of temperature between the drink and the room.

We can model that in this way

\frac{dT}{dt}=k*(T_r-T)

If we rearrange and integrate

\int\frac{dT}{(T-Tr)} =-k*\int dt\\\\ln(T-T_r)=-kt+C1\\\\T-T_r=Ce^{-kt}\\\\T=T_r+Ce^{-kt}

We know that at time 0, the temperature of the drink was 52°F. Then we have:

T=T_r+Ce^{-kt}\\\\52=72+Ce^0=72+C\\\\C=-20

We also know that at t=2, T=55°F

T=T_r+Ce^{-kt}\\\\55=72-20e^{-k*2}\\\\e^{-k*2}=(72-55)/20=0.85\\\\-2k=ln(0.85)=-0.1625\\\\k=0.08

The model for the temperature of the drink can be written as

T=72-20e^{-0.08t}

7 0
4 years ago
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