What is the answer? (384 ft is wrong)

Final answer:

A square cannot be a cross section of a traditional circular cylinder when sliced perpendicular to its height. However, circles, rectangles, and triangles can be cross sections depending on the shape of the cylinder and the slicing.

Explanation:

When considering the cross sections of a cylinder, it is important to remember that the cross-sectional shape remains consistent along the height of the cylinder. A cross section made perpendicular to the height of a traditional circular cylinder is always a circle. Therefore, the correct answer to which shape could not be a cross section of a cylinder is a square, option b. Circles, rectangles, and triangles can all be cross sections of a cylinder depending on the shapes of the top and bottom faces of the cylinder and how the cross section is sliced.

If the cylinder is a right circular cylinder, the only possible perpendicular cross-sectional shape is a circle. A square cross section can occur in a cylinder if the top and bottom faces are square and the cylinder is not circular. Similarly, a rectangular cross section can occur if the cylinder has a rectangular type of cross-section to start with. However, for a circular cylinder, a triangle cannot be a cross section because slicing it will always yield a circular shape due to the round nature of its base.

Answer: A. 3V/ h units^2

Step-by-step explanation:

The slant height is 5 m

The measure of the angle m∠FLD is 45 degrees.

A parallelogram is a special type of quadrilateral that has equal and parallel opposite sides. The given figure shows a parallelogram ABCD which as AB parallel to CD and AD parallel to BC.

In a parallelogram, opposite angles are equal and since it is a quadrilateral, the angles add up to 360°.

[tex]\rm \angle NKL = \angle LMN = X\\\\\angle KLM = \angle KNM = 3X\\\\ 2(x+3x) = 360\\\\ 80x = 360\\\\ x=\dfrac{360}{80}\\\\ x = 45[/tex]

The figure LFND is another quadrilateral where:

[tex]\rm \angle LFN = \angle NDL =90\\\\ \angle FND = 3x[/tex]

The measure of the, m∠FLD is;

[tex]\rm (90\times 2)+3x+ (angle\ FLD) = 360\\\\m \angle\ FLD = 360- 180- (3\times 45)\\\\ m \angle\ FLD = 360 - 315\\\\ m \angle\ FLD = 45[/tex]

Hence, the measure of the angle m∠FLD is 45 degrees.

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The representation of this situation as an inequality is 1.15x ≤ 50. To find the maximum possible cost of Kristin's meal before the tip, the inequality is solved for x, which results in the value of $43.48.

In this scenario, the cost of Kristin's meal before the tip can be represented by the variable x. The tip is 15% of the cost of the meal, which can be represented as 0.15x. The total cost of the meal, including the tip, is therefore x + 0.15x, which is 1.15x. Given that the total cost cannot exceed $50, we can write the following inequality to represent this situation: 1.15x ≤ 50.

To find the most that her meal can cost before a tip, we need to solve this inequality for x. So, we divide both sides by 1.15: x ≤ 50/1.15. After calculation, we find that x (the cost of the meal before the tip) must be ≤ about $43.48.

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The standard deviation for the given test scores can be calculated by finding the mean, computing the squared differences from the mean, averaging those squared differences to get the variance, and then taking the square root of that variance. The standard deviation for the scores is approximately 2.449.

The student is likely seeking assistance with the concept of standard deviation, which is a measure of dispersion or distribution around the mean in a set of data. The question at hand requires calculating the standard deviation for a small dataset consisting of three scores from a chemistry seminar. To find the standard deviation, one would first need to calculate the mean of the test scores, then find the variance by averaging the squared differences between each score and the mean, and finally, take the square root of the variance.

Step-by-step Calculation:

The standard deviation of the test scores in Prof. Miller's advanced chemistry seminar is approximately 2.449.

{ ( -1 , 3 ) , ( 0,4 ), ( 1 , 14 ) , ( 5, 6 ) , ( 7, 2 )}

Function is a relation which each member of the domain is mapped onto exactly one member of the codomain.

There are many types of functions in mathematics such as :

If function f : x → y , then inverse function f⁻¹ : y → x

Let us now tackle the problem!

According to the definition above, it can be concluded that a function cannot have the same x value.

Of the four tables available in choices, table option C has an inverse that is also a function. This is because x values and y values are all different.

[tex]\{(-1,3) , ( 0,4} ) , ( 1,14 ) , ( 5, 6) , ( 7, 2) \}[/tex]

Option A doesn't have inverse because there is the same value of y i.e 4

[tex]\{(-1,-2) , ( 0,\boxed{4} ) , ( 1,3 ) , ( 5, 14) , ( 7, \boxed {4}) \}[/tex]

Option B doesn't have inverse because there is the same value of y i.e 4

[tex]\{(-1,-2) , ( 0,\boxed{4} ) , ( 1,5 ) , ( 5, \boxed {4}) , ( 7, 2) \}[/tex]

Option D doesn't have inverse because there is the same value of y i.e 4

[tex]\{(-1,\boxed {4}) , ( 0,\boxed{4} ) , ( 1,2 ) , ( 5, 3) , ( 7, 1) \}[/tex]

Grade: High School

Subject: Mathematics

Chapter: Function

Keywords: Function , Trigonometric , Linear , Quadratic

Answer:

C.

0.0316

Step-by-step explanation:

A generalization is a broad statement or an idea that is applied to a group of people or things. Most often, generalizations are not entirely true, because usually, the pattern they see in the relationship of a certain group of individuals or situations does not apply. In our example, where Elisa, Sandy, and Jared got sick and were absent from school for  two days straight, within the 3 week period, we can make the following generalizations patterns:

1)      There is an outbreak of sickness in the school for 3 consecutive weeks.

2)      The said sickness lasts for two days.

The best approximation for the volume of the cone in cubic units is 1361.357 units³.

Volume of a three dimensional shape is the space occupied by the shape.

Given that,

Diameter of the cone = 20

Radius of the cone, r = Half of diameter

                                = 20/2 = 10

Height of the cone, h = 13

Volume of the cone = 1/3 π r² h

                                  = 1/3 π (10)² (13)

                                  = 1361.357 units³

Hence the volume is 1361.357 units³.

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A suitable calculator can show you the number of students having a score in that range is expected to be about 57.

To find the interval of 68% of the data, use the empirical rule. The probability a man weighs more than 170 lbs or less than 128 lbs can be found using the normal distribution curve.

To find the interval in which 68% of the data lies, we can use the concept of the empirical rule. According to the empirical rule, approximately 68% of the data falls within one standard deviation of the mean.

Since the standard deviation is 14 pounds and the mean is 142 pounds, one standard deviation above the mean is 142 + 14 = 156 pounds, and one standard deviation below the mean is 142 - 14 = 128 pounds.

Therefore, the interval in which 68% of the data lies is from 128 pounds to 156 pounds.

The probability that a man picked at random from the unit will weigh more than 170 pounds can be found by calculating the area under the normal distribution curve to the right of 170 pounds.

The probability that a man picked at random from the unit will weigh less than 128 pounds can be found by calculating the area under the normal distribution curve to the left of 128 pounds.

Answer:

The answer is B

Step-by-step explanation:

Answer:

[tex]\frac{2m^2n^8}{3}[/tex]

Step-by-step explanation:

The given expression is:

[tex](\frac{4m^{-2}n^8}{9m^{-6}n^{-8}  })^{\frac{1}{2}}[/tex]

Upon simplifying the above equation, we get

Move [tex]m^{-6}[/tex] to teh numerator and Using the negative exponent rule, we get

=[tex](\frac{4m^{-2}n^8m^6}{9n^{-8}  })^{\frac{1}{2}}[/tex]

Move [tex]n^{-8}[/tex] to teh numerator and Using the negative exponent rule, we get

=[tex](\frac{4m^{-2}n^8m^6n^8}{9})^{\frac{1}{2}}[/tex]

=[tex](\frac{4m^{4}n^{16}}{9})^{\frac{1}{2}}[/tex]

=[tex](\frac{(2)^2(m^2)^2(n^8)^2}{(3)^2})^{\frac{1}{2}}[/tex]

=[tex]\frac{2m^2n^8}{3}[/tex]

which is the correct simplified form of the given expression.

a)

The robot made the turn at the point (-20, 15).

b)

The equation of the second line on which the robot traveled is:

y = (4/3)x + 35.

c)

The distance between the starting position and the final position is 20.51 feet.

d)

The equation of the line the robot needs to travel along in order to return and recharge is y = (-3/4)x + 5.

The equation of a line is given by:

y = mx + c

where m is the slope of the line and c is the y-intercept.

Example:

The slope of the line y = 2x + 3 is 2.

The slope of a line that passes through (1, 2) and (2, 3) is 1.

We have,

a.

To find the coordinates of the point at which the robot made the turn, we need to determine its position after traveling for 60 seconds along the initial northeast path.

The distance the robot covers in 60 seconds is:

distance

= speed x time

= (1/2) x 60 = 30 feet

The initial path has a slope of -3/4 (since it passes through (4, -3) and

(0, 5)), so the robot's new path after turning right will have a slope of the negative reciprocal of -3/4, which is 4/3.

We know that the robot traveled 30 feet along the initial path, so we can use the slope and distance traveled to find the endpoint of the initial path:

y = mx + b, where m is the slope and b is the y-intercept

-3 = (-3/4)(4) + b

b = 0

So the equation of the initial path is y = (-3/4)x.

The endpoint of the initial path is (-20, 15) (since the initial path passes through (0, 5) and (-20, 15)).

The robot made the turn at the point (-20, 15).

b.

Since the robot turned right, its new path will be perpendicular to the initial path.

We know the robot's position after turning (i.e., the endpoint of the initial path), so we can use this point and the slope of the new path to find the equation of the new path.

The slope of the new path is 4/3, so the equation of the new path is:

y - 15 = (4/3)(x + 20)

Simplifying this equation, we get:

y = (4/3)x + 35

The equation of the second line on which the robot traveled is:

y = (4/3)x + 35.

c.

From the last equation in part (b), we have:

9(y - 5)² = 29 - 16x² - 12x

Taking the square root of both sides and solving for y, we get:

y = 5 ± √((29 - 16x^2 - 12x)/9)

Since the robot is traveling along the new path for 80 seconds, we want to find y when t = 80.

Substituting t = 80 into the equation for y and simplifying, we get:

y = 5 + √(17/9)

So the robot's final position is approximately (20.50, 6.63).

The distance between the starting position and the final position is:

distance

= √((20.50 - 0)² + (6.63 - 5)²)

= √(421.17)

= 20.51 feet.

d.

The robot needs to travel along a path that is perpendicular to the second line on which it traveled in order to return to its starting position.

The slope of the second line is 4/3, so the slope of the path back to the starting position is -3/4 (the negative reciprocal of 4/3).

We know the robot's final position and the slope of the desired path, so we can use the point-slope form to find the equation of the path:

y - 5 = (-3/4)(x - 0)

Simplifying this equation, we get:

y = (-3/4)x + 5

So the equation of the line the robot needs to travel along in order to return and recharge is y = (-3/4)x + 5.

Setting the two equations equal to each other and solving for x, we get:

(4/3)x + 35 = (-3/4)x + 5

Multiplying both sides by 12 to eliminate fractions, we get:

16x + 420 = -9x + 60

Solving for x, we get:

x = -16.8

Substituting this value of x into either equation, we get:

y = (4/3)(-16.8) + 35

y = 13.60

So the intersection point is approximately (-16.8, 13.60).

To find how long it will take the robot to get there, we can use the distance formula between the robot's final position and the intersection point:

distance = √((-16.8 - 20.50)² + (13.60 - 6.63)²) ≈ 40.49 feet.

Since the robot travels at a constant speed of 1/2 foot per second, it will take approximately 80.98 seconds (rounded to two decimal places) for the robot to return and recharge.

Thus,

The robot made the turn at the point (-20, 15).

The equation of the second line on which the robot traveled is:

y = (4/3)x + 35.

The distance between the starting position and the final position is 20.51 feet.

The equation of the line the robot needs to travel along in order to return and recharge is y = (-3/4)x + 5.

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