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deff fn
15 days ago
15

answer to three friends each create 4 bags of starter bread dough. after ten days, each of those four bags is then divided into

four more bags of dough. how many days pass before 192 bags of dough are created
Mathematics
1 answer:
tester [3.9K]15 days ago
7 0
The solution is 20 days, found as follows:
Each of the three friends initially makes 4 bags of dough. After 10 days, each of these bags is split into 4 new bags:
3 friends × 4 bags = 12 bags, and 12 bags × 4 = 48 bags
Then in the following 10 days, 48 bags are again divided into 4 bags each:
48 × 4 = 192 bags
Adding the periods: 10 days + 10 days = 20 days
Therefore, 192 bags are created after 20 days.
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Water is poured into a conical paper cup at the rate of 3/2 in3/sec (similar to Example 4 in Section 3.7). If the cup is 6 inche
tester [3938]

Response:

The height of the water when it reaches 4 inches is \frac{3}{8\times \pi} inch/s.

Detailed Explanation:

Flow rate of water from the cone = R=\frac{3}{2} inch^3/s

Height of the cup = h = 6 inches

Radius of the cup = r = 3 inches

\frac{r}{h}=\frac{3 inch}{6 inch}=\frac{1}{2}

r = h/2

Volume of the cone = V=\frac{1}{3}\pi r^2h

V=\frac{1}{3}\pi r^2h

\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi r^2h)}{dt}

\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi (\frac{h}{2})^2h)}{dt}

\frac{dV}{dt}=\frac{1}{3\times 4}\pi \times \frac{d(h^3)}{dt}

\frac{dV}{dt}=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}

\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}

h = 4 inches

\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3\times (4inches )^2\times \frac{dh}{dt}

\frac{3}{2} inch^3/s=\pi\times 4\times \frac{dh}{dt} inches^2

\frac{dh}{dt}=\frac{3}{8\times \pi} inch/s

The height of the water when it is 4 inches deep is \frac{3}{8\times \pi} inch/s.

6 0
11 days ago
From a full 50-liter container of a 40% concentration of acid, x liters are removed and replaced with 100% acid. (A) Write the a
tester [3938]

Answer:

A) 20+0.6x

B) Domain is [0, 50] (inclusive)

C) 8.33 litres

Step-by-step explanation:

Assuming there is a 40% acid concentration in a 50-litre container.

The quantity of acid in this container amounts to 40% of 50 litres.

Quantity of acid = \frac{40}{100} \times 50 = 20\ litre

x litres are extracted.

The acid volume removed equals 40% of x litre.

Remaining acid in the container thus becomes = (20 - 40% of x) litre

This is then substituted with pure acid.

Final acid content in the container is therefore = (20 - 40% of x + 100% of x) litre

Quantity of acid in the final mixture:

20 - \dfrac{40}{100} \times x + \dfrac{100}{100} \times x\\\Rightarrow 20 +\dfrac{100-40}{100}x\\\Rightarrow 20 +\dfrac{60}{100}x

Answer A) Final mixture acid content = 20+0.6x

Answer B) x cannot exceed 50 litres (the container's initial volume) and must be at least 0 litres.

Thus, the domain is [0, 50] (inclusive)

Answer C)

For the final mixture to contain 50% acid.

The acid volume = 50% of 50 litres = 25 litres

By applying the equation:

20+0.6x =25\\\Rightarrow 0.6x =5\\\Rightarrow \bold{x =8.33\ litres}

4 0
9 days ago
A 63 liter mixture contains milk and water in a ratio of 4:5. then x liters of milk and y liters of water are added to the mixtu
babunello [3666]

Response:

X+y=237Litres

Detailed explanation:

Let a be the mixture of milk and water.

Define x = milk

Define y = water

Let z = x+y

Total volume of mixture =63litres + z

5/12(3+z))+60=8/15(63-z)

Thus, z =x+y= 237litres

4 0
18 days ago
A chess player moves a knight from the location (3, 2) to (5, 1) on a chessboard. If the bottom-left square is labeled (1, 1), t
lawyer [4039]
To determine the knight's moves from (3, 2) to (5, 1), and knowing the bottom-left corner is (1, 1):
First, the knight moves 2 squares to the left and 1 square down.
Next, to move from (5, 1) to (6, 3), it travels 2 squares to the right, 2 squares up, then 1 square to the left.

5 0
15 days ago
Read 2 more answers
Which statements are true about the graph of y ≤ 3x + 1 and y ≥ –x + 2? Check all that apply. 1.The slope of one boundary line i
Leona [4187]

Answer:

2.Both boundary lines are solid.

3.A solution to the system is (1, 3).

5.The boundary lines intersect.

Step-by-step explanation:

We have

y\leq 3x+1 ----> inequality A

The solution area for inequality A is the region shaded beneath the solid line y=3x+1.

The slope of that solid line is 3.

The coordinates (1,3) fit within the solution set of inequality A.

y\geq -x+2 ----> inequality B

The solution area for inequality B is the shaded part above the solid line y=-x+2.

The slope of this solid line is -1.

The coordinates (1,3) also lie within the solution set of inequality B.

The overall solution for the inequalities is the region shaded between the two solid lines.

Please refer to the attached figure.

Verify each statement

1.The slope of one boundary line is 2.

This statement is False.

2.Both boundary lines are solid.

This statement is True.

3.A solution to the system is (1, 3).

This statement is True.

4.Both inequalities shade below their respective boundary lines.

This statement is False.

5.The boundary lines do intersect.

This statement is True.

The point of intersection is (0.25,1.75).

Refer to the attached figure.

7 0
12 days ago
Read 2 more answers
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