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2 years ago

6

If the ice shop has only enough cocoa to make 4 gallons of chocolate ice cream, how many gallons of strawberry ice cream can the

shop efficiently produce? 0 2 8 12

1 answer:

tankabanditka [31]2 years ago

3 0

The complete questions is:

The ice cream shop needs 5 pounds of strawberries for every gallon of strawberry ice cream. the shop chooses to produce 4 gallons of chocolate ice cream. how many pounds of strawberries should the shop purchase?

Pounds of strawberry ice cream needed to be bought by the shop exists 20 pounds.

<h3>How many gallons of strawberry ice cream can the shop efficiently produce?</h3>

The ice cream shop requires 5 pounds of strawberries for every gallon of strawberry ice cream and the shop chooses to make 4 gallons of ice cream,

Quantity of chocolate ice cream created in the shop = 4 gallons

Quantity of strawberries needed for each gallon of ice cream = 5 pounds

Pounds of strawberries needed to be bought by the shop = Quantity of chocolate ice cream created by the shop $*$ Quantity of strawberries needed for each gallon of ice cream

$$= 4 * 5$

= 20 pounds.

Pounds of strawberries needed to be bought by the shop exists 20 pounds.

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Therefore $k= \frac{ln2 }{18}$ , A=184

Step-by-step explanation:

Given function is

$T(t)=230 -e^{-kt}$

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She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

$138=230 -Ae^{-k\times 18}$

$\Rightarrow -Ae^{-18k}=138-230$

$\Rightarrow Ae^{-18k}=92$ .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

$184=230-Ae^{-k\times 36}$

$\Rightarrow Ae^{-36k}=230-184$

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Dividing (2) by (1)

$\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}$

$\Rightarrow e^{-18k}=\frac{46}{92}$

Taking ln both sides

$ln e^{-18k}=ln\frac{46}{92}$

$\Rightarrow -18k =ln (\frac12)$

$\Rightarrow -18k= ln1-ln2$

$\Rightarrow k= \frac{ln2 }{18}$

Putting the value k in equation (1)

$Ae^{-18\frac{ln2}{18}}=92$

$\Rightarrow A e^{ln2^{-1}}=92$

$\Rightarrow A.2^{-1}=92$

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$\Rightarrow A= 92 \times 2$

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Therefore $k= \frac{ln2 }{18}$ , A=184

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