which equation can be used to represent the distance d for the times given in the table rebeka chose A as the the correct answer.how did she get that answer

Answer:

20+4x

Step-by-step explanation:

4*5=20

4*x=4x

The area enclosed is 3 unit²

the line y=0 simply denotes the x axis. Now, the straight line y= (1/2)x+1 can be written as (x/(-2))+(y/1) = 1. So, this line touches x axis at (-2,0) and y axis at (0,1). the other line y= -(1/4)x+1 can be written as (x/4)+(y/1)=1. So, this line touches y axis at (4,0) and y axis at (0,1). so, it is evident that the triangle that got forms have the vertices of (4,0), (-2,0) and (0,1), the line joining (4,0) and (-2,0) being the base of the triangle.

so, the base of the triangle is 4+2= 6 unit and height of the triangle is 1 unit.

area of the triangle is 1/2* base* height= 1/2*6*1 = 3 unit^2.

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It is true that v1 + 2v2, v2 + 2v3, v4 is also a basis for V.

Suppose that v1, v2, v3, v4 is a basis for vector space V. A basis for a vector space is a set of vectors that are linearly independent and span the entire vector space. In other words, any vector in the vector space can be written as a linear combination of the basis vectors.

In this case, the vectors v1 + 2v2, v2 + 2v3, and v4 are linear combinations of the original basis vectors v1, v2, v3, and v4. Therefore, they are also linearly independent and span the entire vector space V. As a result, the set {v1 + 2v2, v2 + 2v3, v4} is also a basis for V.

In general, any set of vectors that are linearly independent and span the entire vector space can be considered a basis for that vector space. Therefore, it is true that v1 + 2v2, v2 + 2v3, v4 is also a basis for V.

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The answer of the expression is (9c^(3)+79c^(2)+82c+369)/(8(c+7)(c+8)).

To add the rational expressions, we need to find a common denominator. The denominators of the two expressions are (c^(2)+15c+56) and (8c^(2)+120c+448), which can be factored into (c+7)(c+8) and 8(c+7)(c+7), respectively. The common denominator is then 8(c+7)(c+8).

Next, we need to multiply each expression by the appropriate factor to get the common denominator. The first expression is multiplied by 8(c+8)/(8(c+8)) and the second expression is multiplied by (c+7)/(c+7). This gives us:

(8(c+8)(c^(2)-9))/(8(c+7)(c+8)) + ((c+7)(c^(2)+16c+63))/(8(c+7)(c+8))

Simplifying the numerators gives us:

(8c^(3)+56c^(2)-72c-72)/(8(c+7)(c+8)) + (c^(3)+23c^(2)+154c+441)/(8(c+7)(c+8))

Combining the numerators and simplifying gives us:

(9c^(3)+79c^(2)+82c+369)/(8(c+7)(c+8))

This is the fully factored form of the sum of the rational expressions.

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Step-by-step explanation:

a) To form a committee of 8 with 3 prefects, we need to choose 3 prefects from the 5 available prefects, and 5 non-prefects from the remaining 7 boys. We can do this by using the combination formula:

Number of ways = (Number of ways to choose 3 prefects) × (Number of ways to choose 5 non-prefects)

Number of ways to choose 3 prefects from 5 = C(5, 3) = 10

Number of ways to choose 5 non-prefects from 7 = C(7, 5) = 21

Therefore, the total number of committees of 8 with 3 prefects that can be formed is:

Number of ways = 10 × 21 = 210

b) To form a committee of 8 with at least 3 prefects, we need to consider two cases: one where we choose exactly 3 prefects, and one where we choose all 5 prefects. We can calculate the number of ways for each case using the combination formula:

Number of ways to choose exactly 3 prefects and 5 non-prefects = C(5, 3) × C(7, 5) = 210

Number of ways to choose all 5 prefects and 3 non-prefects = C(5, 5) × C(7, 3) = 35

Therefore, the total number of committees of 8 with at least 3 prefects that can be formed is:

Number of ways = (Number of ways to choose exactly 3 prefects and 5 non-prefects) + (Number of ways to choose all 5 prefects and 3 non-prefects)

Number of ways = 210 + 35 = 245

Answer:

b.  6

Step-by-step explanation:

The equation x - 4 = 2 tells us that if we subtract 4 from a number (x), we'll have 2.  Let's add 4 to both sides of the equation:

x - 4 = 2

  +4   +4

    x  = 6

X is 6

Answer:

To find the interquartile range (IQR) of the given scores, we first need to find the median of the dataset.

Finding the Median:

Arranging the scores in order, we get:

57, 66, 81, 97, 121, 133, 146

The median is the middle score. Since there are seven scores, the middle score is the fourth score, which is 97.

Finding the First Quartile:

The first quartile (Q1) is the median of the lower half of the dataset. In this case, the lower half of the dataset consists of the three scores 57, 66, and 81.

Arranging these scores in order, we get:

57, 66, 81

The median of this lower half is (66 + 81) / 2 = 73.5. This is the value of the first quartile, Q1.

Finding the Third Quartile:

The third quartile (Q3) is the median of the upper half of the dataset. In this case, the upper half of the dataset consists of the three scores 121, 133, and 146.

Arranging these scores in order, we get:

121, 133, 146

The median of this upper half is (133 + 121) / 2 = 127. This is the value of the third quartile, Q3.

Finding the Interquartile Range (IQR):

The interquartile range is the difference between the third quartile and the first quartile, so:

IQR = Q3 - Q1 = 127 - 73.5 = 53.5

Therefore, the interquartile range of the scores is 53.5

Step-by-step explanation:

Find the median of the dataset.

To find the median, we first need to arrange the scores in order from smallest to largest:

57, 66, 81, 97, 121, 133, 146

Since there are seven scores in the dataset, the median is the fourth score, which is 97.

Find the first quartile (Q1).

The first quartile is the median of the lower half of the dataset. To find the lower half of the dataset, we need to look at the scores that are less than or equal to the median. In this case, the lower half of the dataset consists of the scores 57, 66, and 81.

To find the median of the lower half of the dataset, we add the two middle scores together and divide by 2. So:

(66 + 81) / 2 = 73.5

Therefore, Q1 = 73.5.

Find the third quartile (Q3).

The third quartile is the median of the upper half of the dataset. To find the upper half of the dataset, we need to look at the scores that are greater than or equal to the median. In this case, the upper half of the dataset consists of the scores 121, 133, and 146.

To find the median of the upper half of the dataset, we add the two middle scores together and divide by 2. So:

(133 + 121) / 2 = 127

Therefore, Q3 = 127.

Find the interquartile range (IQR).

The interquartile range is the difference between Q3 and Q1. So:

IQR = Q3 - Q1 = 127 - 73.5 = 53.5

Therefore, the interquartile range of the scores is 53.5

The interquartile range (IQR) of the scores is 66.

To find the interquartile range (IQR), we first need to find the first quartile (Q1) and the third quartile (Q3).

Step 1: Put the scores in order from smallest to largest:

57, 66, 81, 97, 121, 133, 146

Step 2: Find the median (middle value) of the entire dataset:

The median is the middle value, which is 97 in this case.

Step 3: Divide the dataset into two halves:

The lower half of the dataset includes the scores 57, 66, 81, and 97.

The upper half of the dataset includes the scores 121, 133, and 146.

Step 4: Find the median (middle value) of each half:

The median of the lower half is (66+81)/2 = 73.5.

The median of the upper half is (133+146)/2 = 139.5.

Step 5: Find the interquartile range (IQR):

IQR = Q3 - Q1

Q1 = 73.5 and Q3 = 139.5

IQR = 139.5 - 73.5 = 66.

Therefore, the interquartile range (IQR) of the scores is 66.

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The solution for v in the equation (4)/(5v)=(1)/(10)-(3)/(2v) is v = 0.

To solve for v in the equation (4)/(5v)=(1)/(10)-(3)/(2v), we need to use the following steps:

Step 1: Multiply both sides of the equation by the common denominator of 10v to get rid of the fractions. This gives us:
(4)(10v)/(5v) = (1)(10v)/(10) - (3)(10v)/(2v)

Step 2: Simplify the equation by canceling out the common terms. This gives us:
8v = v - 15v

Step 3: Combine like terms on the right side of the equation. This gives us:
8v = -14v

Step 4: Add 14v to both sides of the equation to isolate v on one side. This gives us:
22v = 0

Step 5: Divide both sides of the equation by 22 to solve for v. This gives us:
v = 0/22

Step 6: Simplify the fraction to get the final answer. This gives us:
v = 0

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The solution set of the inequality x-15≤6/9 is  x≤47/3.

a relationship between two expressions or values that are not equal to each other is called 'inequality.

The given inequality is x-15≤6/9

x minus fifteen lesser than or equal to six by nine.

x is the variable and minus is the operator,

x-15≤6/9

On the right side the numerator and denominator is divided by 3

x-15≤2/3

Add 15 on both sides

x≤2/3+15

x less than or equal to two by three plus fifteen.

x≤2+3(15)/3

When three is multiplied with fifteen we get 45.

x≤45+2/3

So when forty five and two are added we get 47.

x≤47/3

Hence, the solution set of the inequality x-15≤6/9 is  x≤47/3.

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Complete quesstion,

Find the solution set of the  inequality x-15≤6/9

The ordered pair that is a member of both equations is (-27/7, -11/7).

To find the ordered pair that is a member of both equations, we need to solve the system of equations. We can do this by using the elimination method.

First, we will multiply the first equation by 2 and the second equation by -4 to eliminate the x variable:

-4x - 8y = 28
-20y - 8x = 16

Next, we will add the two equations together:

-28y = 44

Now we can solve for y:

y = -44/28
y = -11/7

Now we can plug this value of y back into one of the original equations to solve for x:

-2x - 4(-11/7) = 14
-2x + 44/7 = 14
-2x = 14 - 44/7
-2x = 54/7
x = -27/7

Therefore, the ordered pair that is a member of both equations is (-27/7, -11/7).

Answer: The ordered pair that is a member of both equations is (-27/7, -11/7).

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Answer:

We can use a recursive approach to solve this problem. Let's define a function $f(n)$ as the number of ways to ascend a flight of $n$ steps, where the priestess can only take one or two steps at a time.

To ascend a flight of 1 step, there is only one way to do it (take one step). Therefore, $f(1) = 1$.

To ascend a flight of 2 steps, there are two ways to do it (take two steps at once or take one step twice). Therefore, $f(2) = 2$.

For a flight of $n$ steps, the priestess can either take one step and ascend the remaining $n-1$ steps in $f(n-1)$ ways, or take two steps and ascend the remaining $n-2$ steps in $f(n-2)$ ways. Therefore, we have the recursive formula:

$$f(n) = f(n-1) + f(n-2)$$

Using this formula, we can calculate the number of ways to ascend a flight of 5 steps:

$$f(1) = 1$$

$$f(2) = 2$$

$$f(3) = f(2) + f(1) = 3$$

$$f(4) = f(3) + f(2) = 5$$

$$f(5) = f(4) + f(3) = 8$$

Therefore, there are 8 different ways for the priestess to ascend the steps. To find out if she can do this a different way each day for a week, we need to make sure that there are at least 7 different ways. Since there are 8 ways, it is possible for the priestess to ascend the steps in a different way each day for a week.

The absolute deviation for 612 in the given data set is 14.8.

The average distance between each data value and the mean is known as the mean absolute deviation (MAD) of a data set.

A measure of variance in a data set is the mean absolute deviation.

We may determine how "spread out" the values in a data collection are by looking at the mean absolute deviation.

So, get the mean as follows:
80+75+25+62+68)/5

310/5

62

Now, get the distance of each number as follows:
-----------------

| 80     |  18  |

|   75   |   13 |

|+   25   |   37 |

|    62  |    0  |

|   68   |   6   |

-----------------

Now, get the distance:

18+13+37+0+6/5

14.8

Therefore, the absolute deviation for 612 in the given data set is 14.8.

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What is the absolute deviation for 62 in the data set?

{80, 75, 25, 62, 68}

The required cost of the cans of weedkiller needed to cover lawn are  £ 47.4.

The formula A = 12 (a Plus b) h is used to calculate the area of a trapezoid, where a and b are the bases (parallel sides) and h is the height (the angle between the bases).

We have;

Dimensions of Trapezoid; a = 20, b = 12, Height = h

So;

[tex]$Area = \frac{(a+b)h}{2}$[/tex]

For h, Using Pythagoras theorem;

17² = 8² + h²

h = 15 m

Then,

[tex]$Area = \frac{(12+20)15}{2}$[/tex]

Area = 240 Sq meter

Then,

100  Sq meter = £19.75

For 240  Sq meter

= £19.75(2.4)

= £ 47.4

Thus, required cost is  £ 47.4.

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The equation of exponential function=y = 0.054 × [tex]1.028^{x}[/tex]

As the name implies, exponents are utilised in exponential functions. But remember that a variable serves as the exponent instead of a constant as the basis of an exponential function.

A function is a power function rather than an exponential function if its base is a variable and its exponent is a constant.

If the graph of exponential function passes through the points (1960,19.005) and (2010,19.548), then the coordinates of these points satisfy the equation, y = [tex]ab^{x}[/tex]

Now, 19.005 = a × [tex]b^{1960}[/tex]

⇒  [tex]b^{1960}[/tex] = 19.005/a

Similarly, 19.548 = a × [tex]b^{2010}[/tex]

⇒ [tex]b^{2010}[/tex] = 19.548/a

⇒  [tex]b^{1960}[/tex] × [tex]b^{50}[/tex] = 19.548/a

⇒ [tex]b^{50}[/tex] = 19.548/a × a/19.005

         = 1.028

Putting this value to find a,

we get a = 0.054

So, the equation of exponential function=

y = 0.054 × [tex]1.028^{x}[/tex]

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The value of [tex]x[/tex] that satisfies the equation is [tex]x = 6[/tex].

How can we solve an equation with a fraction and a quadratic expression?

To solve an equation with a fraction and a quadratic expression, we can first try to simplify the fractions by finding a common denominator. Then, we can move all the terms to one side of the equation and simplify it into a quadratic equation. Finally, we can solve the quadratic equation using factoring or the quadratic formula.

Find the value of [tex]x[/tex]:

The given equation is:

[tex]\frac{50}{4x-12}+\frac{x-4}{x^2+x-12}=\frac{x}{2}[/tex]

The first step is to find the LCM of the denominators. Factoring the denominator [tex]x^2+x-12[/tex], we get:

[tex]x^2+x-12=(x+4)(x-3)[/tex]

So the LCM of the denominators is [tex](4x-12)(x+4)(x-3)[/tex]. Multiplying both sides of the equation by this LCM, we get:

[tex]50(x+4)(x-3)+(x-4)(4x-12)=\frac{x}{2}(4x-12)(x+4)(x-3)[/tex]

Expanding and simplifying both sides, we get:

[tex]6x^3-26x^2-13x+56=0[/tex]

Factoring out [tex](x-4)[/tex], we get:

[tex](x-4)(6x^2-10x-14)=0[/tex]

Solving for [tex]x[/tex] using the quadratic formula, we get:

[tex]x=\frac{5\pm\sqrt{29}}{3} \text{ or } x=4[/tex]

However, we need to check if any of these solutions make the denominator zero, which would be undefined. Checking each of the possible solutions, we find that [tex]x=4[/tex] is the only solution that does not make any of the denominators zero.

Therefore, the solution is [tex]x=4[/tex].

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The probability that the person chosen at random is a boy is 8/13 and the probability that they are between 120 cm and 130 cm tall is 6/10.

Polygon is a closed figure that consists of three or more straight lines. It is a two-dimensional shape with straight sides and angles. Polygons are used to construct many shapes in geometry such as triangles, rectangles, squares and many more. Polygons are used in many fields such as computer graphics, navigation, engineering and architecture.

The frequency polygon show that there are more boys than girls in the class. This can be seen from the higher peak on the boys side of the graph. The total number of boys is 8 and the total number of girls is 5. This gives us a probability of 8/13 that the person chosen at random is a boy.

To work out the probability that the person is between 120 cm and 130 cm tall, we must first count the number of individuals in the class in this height range. From the graph, we can see that there are 6 boys and 4 girls in this range. This gives us a probability of 6/10 that the person chosen at random is between 120 cm and 130 cm tall.

In conclusion, the probability that the person chosen at random is a boy is 8/13 and the probability that they are between 120 cm and 130 cm tall is 6/10.

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Ordered pair (0, 3), (1, 48) and many more lies on the graph of the function.

What are Functions?

A function in math is a relation between a set of inputs (called the domain) and a set of possible outputs (called the range) with the property that each input is related to exactly one output. It can be represented using an equation, graph, or table of values.

The exponential function [tex]f(x) = 3(4)^(2x)[/tex] can be evaluated for any value of x to get the corresponding value of y. To find ordered pairs that lie on the graph of this function, we can substitute different values of x into the function and calculate the corresponding values of y.

For example, if we substitute x = 0 into the function, we get:

[tex]f(0) = 3(4)^(2(0)) = 3(4)^0 = 3(1) = 3[/tex]

So the ordered pair (0, 3) lies on the graph of the function.

Similarly, if we substitute x = 1 into the function, we get:

[tex]f(1) = 3(4)^(2(1)) = 3(4)^2 = 3(16) = 48[/tex]

So the ordered pair (1, 48) also lies on the graph of the function.

We can repeat this process for other values of x to find more ordered pairs that lie on the graph of the function.

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In this case, the mean of the data set is 67, the sum of all the data values is 2291, the sum of all the squared data values is 182,361, the sample standard deviation is 8, the population standard deviation is 8.024961059, and the sample size is 29.

The TI-84 Plus display presents sample statistics for a set of data. The "x" symbol represents the mean, or average, of the data set. The "Σx" symbol represents the sum of all the data values, and the "Σx2" symbol represents the sum of all the squared data values.

These statistics can be used to analyze and interpret the data set. For example, the mean can be used to determine the central tendency of the data, and the standard deviation can be used to measure the spread of the data.

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a) The length of DP is,

DP = 11 Units

b) The lengths are,

RY = 12

YN = 24

YT = 10

BT = 30

IY = 10

And, IE = 15

TN = 25

And, IN  = 50

A triangle is a three sided polygon, which has three vertices and three angles which has the sum 180 degrees.

Given that;

Triangle are shown in figure.

a) We have;

AM = MP

Hence,

12x - 4 = 4x + 12

12x - 4x = 4 + 12

8x = 16

x = 2

Hence, Lenght of DP is,

DP = 7x - 3

DP = 7×2 - 3

DP = 14 - 3

DP = 11

b) We know that;

RY : YN = 1 : 2

Here, RN = 36

Hence, We get;

36 = x + 2x

3x = 36

x = 12

Hence,

RY = x = 12

YN = 2x = 2 x 12 = 24

And, If BY = 20;

YT = 20/2

YT = 10

And, BT = 20 + 10 = 30

If EY = 5;

IY = 2 × 5 = 10

And, IE = 5 + 10 = 15

If IT = 25

Then, TN = 25

And, IN = 25 + 25 = 50

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The given function is a piecewise function and we are asked to evaluate f(0). To do this, we need to find the expression that corresponds to x = 0. We can see that x < 2, so the expression we need to use is 1 + x.

Plugging in x = 0 into this expression gives us:

f(0) = 1 + 0 = 1

Therefore, the answer is 1.

In HTML format, the answer would be:

The given function is a piecewise function and we are asked to evaluate f(0). To do this, we need to find the expression that corresponds to x = 0. We can see that x < 2, so the expression we need to use is 1 + x.

Plugging in x = 0 into this expression gives us:

f(0) = 1 + 0 = 1

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137.84 cm to the nearest tenth would be 137.8 cm

1- 20x12= 240

2- 12x32= 384

3- 21x21= 441

Answer:

Step-by-step explanation:

Answer:

[tex] = 3kd + 2d + 5r + 6kr[/tex]

Step-by-step explanation:

[tex]7dk + 6kr + 4d + 8r - 4 kd - 2 d - 3r \\ = 3kd + 2d + 5r + 6kr[/tex]

Answer:

Step-by-step explanation:

Let [tex]l[/tex] be length, [tex]b[/tex] be breadth.

[tex]l=3b[/tex]   [tex](a)[/tex]                (the length is 3times the breadth)

[tex]2l+2b=40[/tex]   [tex](b)[/tex]         (perimeter is 40)

Now we must solve these to find [tex]l[/tex] and [tex]b[/tex]:

Substitute [tex](a)[/tex] into [tex](b)[/tex] :

[tex]2(3b)+2b=40[/tex]

[tex]6b+2b=40[/tex]

[tex]8b=40[/tex]

[tex]b=5cm[/tex]

Substitute [tex]b=5[/tex] into [tex](a)[/tex]:

[tex]l=3\times 5=15cm[/tex]

Find area:

[tex]A=lb=15 \times 3=45cm^2[/tex]

Answer: θ=0.75448...πn