8.Find the value of c that makes the expression be a perfect square trinomial p^2-30p+c

9.Use quadratic formula to solve. 4x^2+7x-15=0

10.Find the number of real number solutions using the discrimination. x^2+4x-32=0

Please help these are my final questions and I need help with them

Answer:

D.3x(x-2)

Step-by-step explanation:

i already did this

Answer:

the cost is [tex]\$6[/tex]

Step-by-step explanation:

observing the graph

we know that

For the interval of x------> [tex](0,1][/tex] -----> the cost is [tex]\$0[/tex]

For the interval of x------> [tex](1,3][/tex] -----> the cost is [tex]\$4[/tex]

For the interval of x------> [tex](3,4][/tex] -----> the cost is [tex]\$6[/tex]

For the interval of x------> [tex](4,infinite)[/tex] -----> the cost is [tex]\$10[/tex]

In this problem we have

[tex]3[/tex] hours and [tex]45[/tex] minutes

therefore

the value of x belong to the interval [tex](3,4][/tex]

the cost is [tex]\$6[/tex]

1/2(z+18) is the expression of 1/2z+9 when 1/2 is factored out.

A fraction represents a part of a whole.

Given,

The expression is 1/2 z+9

A factor is a number that divides another number, leaving no remainder.

The given expression is one by two times of z plus nine.

1/2 z+9

Now we need to take 1/2 as common from the expression

1/2(z+18)

Hence, 1/2(z+18) is the expression of 1/2z+9 when 1/2 is factored out.

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1. The dataset with the smallest IQR is the Mid-term exams. 2. The exam median is only 1–2 points higher than the class median. 3. The standard deviation is the best measurement of spread for games and movies.

1. The dataset with the smallest IQR is the Mid-term exams. To find the IQR, first, calculate the first quartile (Q1) and the third quartile (Q3). Then, find the difference between Q3 and Q1. By comparing the IQR values for the Mid-term and Final exams, it can be determined that the Mid-term exams have the smallest IQR.

2. The exam median is only 1–2 points higher than the class median. The box plot's median represents the middle value of the dataset. By comparing the medians of the exam and class data, it can be determined that the exam median is only 1–2 points higher than the class median.

3. The standard deviation is the best measurement of spread for games and movies. While the IQR can measure spread, the standard deviation is a more precise measurement. Comparing the spread of the data, the standard deviation is the best measurement for both games and movies.

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Various statistical measures such as IQR, Standard Deviation, Median and Mean were used to interpret the given data in numerous scenarios. Outliers were identified and impacts on datasets were evaluated.

To answer these questions, we need to understand few key statistical terms: 'Mean' is the simple average of data, 'Median' is the middle score of data,  'IQR' (Inter Quartile Range) is the difference between the upper quartile (Q3) and the lower quartile (Q1), which helps in understanding the spread and '.'Standard Deviation' measures the absolute variability of a dataset.

Question 1: IQR for the mid-term exams is Q3 (92) - Q1 (82) = 10. IQR for the final exams is Q3 (85) - Q1 (78) = 7. So, the final exams have the smallest IQR.

Question 2: The boxes showing median indicates that the exam median is only 1–2 points higher than the class median.

Question 3: Since the spread of the data at basketball games and local movie theaters demonstrates varied distributions, both the IQR and the standard deviation should be used to evaluate the data spread.

Question 4: Box plots do not provide direct information on the mean. So, there is no way of telling what the means are.

Question 5: For the video hours, we see that values 18, 21, and 26 lie far from the main part of the data, thus they can be considered as possible outliers.

Question 6: The male data set shows a high data value, which causes the data set to be asymmetrical. This could affect statistical measures like the mean and standard deviation.

Question 7: The spread of the college data, having a large standard deviation and IQR, is best described by the standard deviation. The high school data, with smaller numbers, is best described by the IQR.

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Final answer:

To find out if the ratios 1/9 and 6/54 form a proportion, calculate their cross-products. Since both cross-products are equal (54 = 54), they do form a proportion.

Explanation:

The question asks if two given ratios form a proportion. To determine whether ratios form a proportion, you can compare the cross-products of the ratios. If the cross-products are equal, the ratios do form a proportion. For the given ratios 1/9 and 6/54, we will calculate the cross-products:

For the first ratio, 1 times 54 equals 54.

For the second ratio, 9 times 6 equals 54.

Since both cross-products are equal (54 = 54), the ratios 1/9 and 6/54 do indeed form a proportion.

To write a proportion by setting two ratios equal to one another with the unit 'meters', an example would be 1/20 = 1/5.5. When working with unit scales or unit rates, such as in map readings or speed, you compare two measurements where one of the ratios is typically a value of 1, like the example provided 55 miles per hour, which is written as 55/1 miles/hour.

The colon will blink 8400 times in 2 hours and 20 minutes.

Since the colon blinks every second. And we know that there are 60 minutes in an hour and there are 60 seconds in a minute.

Thus, we can say:

1 minutes = 60 seconds

1 hour = 60 minutes = 3600 seconds

Note: 60 minute = (60 * 60) seconds = 3600 seconds

Thus, in 2 hours and 20 minutes, we have:

2 hours + 20 minutes = (2 * 3600) + (20 * 60)

                                   = 7200 + 1200

                                   =  8400 second

Therefore, the colon will blink 8400 times in 2 hours and 20 minutes.

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Answer:

Step-by-step explanation:

It will take 24 minutes to fill the tub when both the faucet is running and the drain is open. The calculation is done using rates and subtraction.

This problem can be approached with the concept of rates. The rate at which the bathtub fills is 1 tub per 8 minutes or 1/8 tubs/min. Similarly, the rate at which it drains is 1 tub per 12 minutes or 1/12 tubs/min. When the faucet is running and the drain is open, the net rate is the fill rate minus the drain rate.

So, (1/8 - 1/12) tubs/minute = 1/24 tubs/minute. Thus, it will take 24 minutes to fill the bathtub with both the faucet running and the drain open.

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Answer:

Step-by-step explanation:

We have to find the values of the given trigonometric ratios at the angle indicated. Thus,

(A) The given trigonometric function is:

[tex]cos270^{\circ}[/tex]

Since, the function lies in Quadrant III, therefore the value of the function will be negative.

Also, [tex]cos270^{\circ}=-(0)=0[/tex]

(B) The given trigonometric function is:

[tex]sin270^{\circ}[/tex]

Since, the function lies in Quadrant III, therefore the value of the function will be negative.

Also, [tex]sin270^{\circ}=-1[/tex]

(C) The given trigonometric function is:

[tex]tan270^{\circ}[/tex]

Since, the function lies in Quadrant III, therefore the value of the function will be negative.

Also, [tex]tan270^{\circ}=undefined[/tex]

(D) The given function is:

[tex]cos0^{\circ}[/tex]

Since, the function lies in Quadrant I, therefore the value of the function will be positive.

Also,  [tex]cos0^{\circ}=1[/tex]

(E) The given function is:

[tex]sin0^{\circ}[/tex]

Since, the function lies in Quadrant I, therefore the value of the function will be positive.

Also,  [tex]sin0^{\circ}=0[/tex]

(F) The given function is:

[tex]tan0^{\circ}[/tex]

Since, the function lies in Quadrant I, therefore the value of the function will be positive.

Also,  [tex]tan0^{\circ}=0[/tex]

A perimeter of 24 feet is achieved in this optimal scenario.

For an area of 36 square feet, we can factor this into pairs: (1, 36), (2, 18), (3, 12), (4, 9), and (6, 6). Since we are looking for the minimum perimeter, the closest factors will yield the smallest perimeter, which in this case is the pair (6, 6). Therefore, a rectangle with the dimensions 6 feet by 6 feet will not only satisfy the area requirement of 36 square feet but will also have the minimum perimeter, which is 24 feet.

The perimeter (P) of a rectangle is calculated by the formula P = 2(l + w), where 'l' is the length and 'w' is the width. Given our dimensions, P = 2(6 + 6) = 24 feet. The square shape of the rectangle, which is a special case where the length and width are equal, is what minimizes the perimeter for a given area.

The rectangle with an area of 36 square feet that has minimal perimeter is actually a square with dimensions of 6 feet by 6 feet.

To find the length and width of a rectangle with an area of 36 square feet that has the minimum perimeter, we need to follow these steps:

Thus, the dimensions that yield the minimum perimeter are 6 feet by 6 feet. This results in a square having the smallest possible perimeter for the given area.

The estimated amount of fabric that the umbrella has is approximately [tex]\(112 + 32\sqrt{2}\)[/tex] square feet.

To estimate the amount of fabric that the umbrella has, we first need to find the total surface area of the regular octagonal pyramid.

A regular octagonal pyramid consists of 8 congruent isosceles triangles as its lateral faces and a regular octagon as its base.

Given:

- Side length of the octagon (s) = 4 feet

- Slant height of the pyramid (l) = 5 feet

First, let's calculate the area of one of the lateral faces (isosceles triangle) using the formula for the area of a triangle:

[tex]\[ \text{Area of one lateral face} = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]

In this case, the base of the triangle is the side of the octagon, s=4 feet, and the height is the slant height of the pyramid, l=5 feet. Therefore:

[tex]\[ \text{Area of one lateral face} = \frac{1}{2} \times 4 \times 5 = 10 \text{ square feet} \][/tex]

Since there are 8 congruent lateral faces on the octagonal pyramid, the total area of the lateral faces is:

[tex]\[ \text{Total area of lateral faces} = 8 \times 10 = 80 \text{ square feet} \][/tex]

Now, let's calculate the area of the base (regular octagon). The formula for the area of a regular octagon is:

[tex]\[ \text{Area of octagon} = 2 \times (1 + \sqrt{2}) \times s^2 \]\[ \text{Area of octagon} = 2 \times (1 + \sqrt{2}) \times 4^2 \]\[ \text{Area of octagon} \approx 2 \times (1 + \sqrt{2}) \times 16 \]\[ \text{Area of octagon} \approx 32 \times (1 + \sqrt{2}) \][/tex]

Now, we can calculate the total surface area of the regular octagonal pyramid by adding the area of the lateral faces and the area of the base:

[tex]\[ \text{Total surface area} = \text{Area of lateral faces} + \text{Area of octagon} \]\[ \text{Total surface area} = 80 + 32 \times (1 + \sqrt{2}) \]\[ \text{Total surface area} \approx 80 + 32 \times (1 + \sqrt{2}) \]\[ \text{Total surface area} \approx 80 + 32 + 32\sqrt{2} \]\[ \text{Total surface area} \approx 112 + 32\sqrt{2} \][/tex]

Given is a circle O with tangent AB and secant OB.

Given is OA = 21 units and BC = 14 units.

From the diagram, OB = OC + BC.

OC and OA, both are radius, so OC = OA = 21 units.

Now OB = 21 + 14 = 35 units.

In right triangle ΔOAB, using Pythagorean theorem;

OA² + AB² = OB²

⇒ (21)² + AB² = (35)²

⇒ 441 + AB² = 1225

⇒ AB² = 1225 - 441  = 784 square units

⇒ AB = [tex]\sqrt{784} =28[/tex]

⇒ AB = 28 units.

Hence, final answer is AB = 28 units.

The playhouse floor is a rectangle with dimensions of 6 by 8.5 feet. To determine the amount of carpet needed, multiply the length and width to calculate the area, which is 51 square feet.

Karl and his dad need to know the amount of carpeting required to cover the floor of a playhouse, which is a practical mathematics problem involving area calculation. To find out how much carpet they need, they have to calculate the area of the rectangular floor, which is the product of its length and width.

The floor measures 6 feet in length and 8.5 feet in width. Multiplying these two dimensions gives us the area:

Area = Length times Width

Area = 6 ft times 8.5 ft

Area = 51 square feet

Therefore, Karl and his dad would need to purchase 51 square feet of carpeting to cover the playhouse floor.

Answer:

A) g(x)=x2−12x+31