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3 years ago

What's the cost of the # of cookies?

Mathematics

2 answers:

77julia77 [94]3 years ago

8 0

You would choose a cost that will increase each time a cookie is added

pochemuha3 years ago

8 0

Try 3 :)))) hope it helps

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To rent a certain meeting room, a college charges a reservation fee of $15 and an additional fee of $6 per hour. The chemistry c

Dafna1 [17]

Answer:

To spend at most $93, they need to rent the room less than or equal 13 hours.

Step-by-step explanation:

<u><em>The complete question is</em></u>

To rent a certain meeting room, a college charges a reservation fee of $15 and an additional fee of $6 per hour. The chemistry club wants to spend at most $93 on renting a room. What are the possible numbers of hours the chemistry club could rent the meeting room? Use t for the number of hours. Write your answer as an inequality solved for t,

Let

t ----> the number of hours

we know that

I this problem the word "at most" means "less than or equal to"

The number of hours rented multiplied by the cost per hour, plus the reservation fee, must be less than or equal to $93

The inequality that represent this situation is

$6t+15\leq 93$

solve for t

subtract 15 both sides

$6t\leq 93-15$

$6t\leq 78$

Divide by 6 both sides

$t\leq 13\ hours$

therefore

To spend at most $93, they need to rent the room less than or equal 13 hours.

6 0

3 years ago

Which function has a graph with a horizontal asymptote at y = 3, a vertical asymptote at x = 1, and an x-intercept at 2?

pickupchik [31]

B. f(x) = 2(4)x–3 (e2020)

7 0

3 years ago

Read 2 more answers

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

Joe bought 5 apples and 4 bananas for $6. Dawn bought 3 apples and 6 bananas for $6.30. How much foes each apple and each banana

zepelin [54]

Answer:

Apple = $0.6

Banana = $0.75

Step-by-step explanation:

Let us represent Apple by A and Banana by B

For Joe,

5A + 4B = 6 (1)

For Dawn

3A + 6B = 6.3 (2)

To eliminate A, Multiply equation (1) by 3 and equation (2) by 5

15A + 12B = 18 (3)

15A + 30B = 31.5 (4)

subtracting equation (4) from (3)

15A - 15A +12B - 30B = 18 - 31.5

-18B = -13.5

divide through by -18

-18B/-18 = -13.5/-18

B = 0.75

substitute 0.75 in equation (1)

5A + 4B = 6

5A + 4(0.75) = 6

5A + 3 = 6

subtract 3 from both sides

5A + 3 - 3 = 6 - 3

5A = 3

divide through by 5

5A/5 = 3/5

A = 0.6

4 0

3 years ago

A store buys 6 sweaters for $30 and sells them for $132. How much product does the store make per sweater

masya89 [10]

Answer:

$102

Step-by-step explanation:

You subtract $30 from $132. That gives you: $102.

I hope this helps, and i hope that it's correct. sorry if its not. have a great day!

4 0

3 years ago