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fomenos
19 days ago
6

A beverage manufacturer performs a taste-test and discovers that people like their fizzy beverages best when the radius of the b

ubbles is about .3 mm. According to the formula below, what would be the volume of one of these bubbles?
A. About 0.27 mm^3
B. About 0.11 mm^3
C. About 0.38 mm^3
D. About 0.42 mm^3

Mathematics
2 answers:
tester [8.8K]19 days ago
8 0

WHEN THE RADIUS REACHES 0.7mm, THE VOLUME IS APPROXIMATELY 1.44mm^3 APEX;)

tester [8.8K]19 days ago
5 0
Since the radius is provided, we can formulate the equation and compute the volume V.

0.3= \sqrt[3]{ \frac{3V}{4 \pi } }

Next, we will cube both sides to eliminate the cube root resulting in:

0.027= \frac{3V}{4 \pi }

Then, to remove the fraction, we multiply both sides by 4 pi:

0.339=3V

Finally, we can divide both sides by 3 to derive V

V=0.11

Thus, the volume amounts to 0.11 cubic mm.


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B refers to the base of the triangle,
and a signifies the length of the two identical sides.


The measurement labeled as 'a' is larger than 'b' since those equal sides are longer than the base. Given "one of the longer sides measures 6.3 cm," we assign a = 6.3.


Substitute 6.3 for each 'a' in the equation and solve for b:
2a + b = 15.7
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1 month ago
Problem 5 (4+4+4=12) We roll two fair 6-sided dice. Each one of the 36 possible outcomes is assumed to be equally likely. 1) Fin
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1 2 3 Step-by-step explanation: Generally, during the roll of two fair 6-sided dice, the doubles result in (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6). Therefore, the total for doubles is N = 6. The outcome of rolling two fair 6-sided dice yields n = 36. Thus, the probability of rolling doubles (matching numbers on both dice) is calculated mathematically. When rolling two fair dice, outcomes that sum to 4 or less are (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1). Observing this, we see two doubles present. Consequently, the conditional probability of rolling doubles is represented mathematically. Lastly, when rolling the two fair dice, outcomes that show different numbers result in L = 30, while outcomes where at least one die shows a 1 give W = 10. Hence, the conditional probability of having at least one die show a 1 is presented mathematically.
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14 days ago
A rectangular schoolyard is to be fenced in using the wall of the school for one side and 150
lawyer [9226]

Answer:

Part 1) The equation is A(x)=150x-2x^2

Part 2) When x=40 m, the area of the schoolyard is A=2,800 m^2

Part 3) The valid domain consists of all real numbers exceeding zero and below 75 meters

Step-by-step explanation:

Part 1) Formulate an expression for A(x)

Let

x -----> the length of the rectangular school yard

y ---> the width of the rectangular school yard

It is known that

The perimeter for the fencing (taking the school wall as one side) is

P=2x+y

P=150\ m

thus

150=2x+y

y=150-2x -----> this is equation A

The area of the rectangular school yard is

A=xy ----> this is equation B

Substituting equation A into equation B yields

A=x(150-2x)

A=150x-2x^2

Change to function notation

A(x)=150x-2x^2

Part 2) What is the area when x=40?

With x equal to 40 m

substitute the value into the expression from Part 1 to determine A

A(40)=150(40)-2(40^2)

A(40)=2,800\ m^2    

Part 3) What would be a suitable domain for A(x) in this scenario?

We understand that

A signifies the area of the rectangular school yard

x characterizes the length of the rectangular school yard

It follows that

A(x)=150x-2x^2

This forms a vertical parabola opening downwards

The vertex indicates a maximum point

The x-coordinate of the vertex corresponds to the length that maximizes the area

The y-coordinate of the vertex denotes the maximum area

The vertex corresponds to (37.5, 2812.5)

Refer to the accompanying figure

Consequently,

The peak area achieved is 2,812.5 m^2

The x-intercepts are located at x=0 m and x=75 m

The domain for A is the range -----> (0, 75)

All real numbers greater than zero and less than 75 meters

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1 month ago
1. Darnell reads at a constant rate
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Answer:

Darnell reads 1,715 words in 7 minutes.

Step by step Explanation:

1. First, determine how many words he can read in a minute by dividing 735 words by 3. The result is 245.

245

______

3)735

6 drop the 3 to form 13

-_____

1 3

12 drop the 5 to make 15

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1 5

15

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0

2. Next, since he reads 735 words over 3 minutes, multiply that by 2 to find words read in 7 minutes: 3×2=6, thus, 735+735 (735×2) equals 1,470 words.

3. Finally, add 245 to account for the last minute we calculated. Therefore, the total is 1,715 words in 7 minutes.

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24 days ago
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