Write an expression that is equivilant to 5(6+y)

Answer:

0.84 miles

Step-by-step explanation:

80% is the same as 0.8, so 0.8*4.2=3.36. This means he has completed 3.36 miles. So, to find how much he has left, subtract. 4.2-3.36=0.84. Therefore, he has 0.84 miles left in his hike.

Answer: 0.84 miles

Step-by-step explanation:

80% is the same as 0.8, so 0.8*4.2=3.36. This means he has completed 3.36 miles. So, to find how much he has left, subtract. 4.2-3.36=0.84. Therefore, he has 0.84 miles left in his hike.

Answer:

c = 12

Step-by-step explanation:

This is a special triangle, also known as the 30-60-90 triangle. See the help image attached for another special triangle known as the 45-45-90 triangle. 30-60-90 triangle has specific rules that we can apply to this problem:

With that being said....

c = 2x

6 = x

a = x√3

We know that x = 6, so now solve for a & c:

c = 2(6) = 12

a = 6√3

Hope this helps!

Answer:

18/1,200?

Step-by-step explanation:

im not too sure on this

Answer:

B. Alternate Interior

Step-by-step explanation:

I hope this helps!!!!!

Step-by-step explanation:

and plot the normal 1 , I am pretty sure

9514 1404 393

Answer:

Step-by-step explanation:

You are given two ordered pairs:

  (years after 1990, cost per credit hour) = (x, C) = (0, 99) and (7, 176)

(Note that 1997 is 7 years after 1990.)

__

The "slope" or "rate of change" of the linear model of this relationship is found from ...

  m = (change in C)/(change in x) = (176 -99)/(7 -0) = 77/7 = 11

Then the linear model will be ...

  C = mx + b . . . . . for some value of b

Using the first ordered pair (x, C) = (0, 99), we can find b:

  99 = 11×0 +b

  99 = b

So, the linear equation for cost of a credit hour can be written ...

  C = 11x +99

__

The year 2000 is 10 years after 1990, so the cost when x=10 is ...

  C = 11×10 +99 = 209

In the year 2000, tuition will be $209 per credit hour.

__

To answer the last question, we need to find x when C = 231. Put this in the equation, and solve:

  231 = 11x +99

  132 = 11x . . . . . . . subtract 99

  132/11 = x = 12 . . . divide by 11

12 years after 1990 is 2002.

In the year 2002, tuition will be $231 per credit hour.

After we evaluate the function f at the indicated values the results are

a. (4) −11, (5) −52, (6) 0

b. (4) -1, (5) −17, (6) -5

c. (4) −2a−5, (5) −5a² −2a−1, (6) −3+a

d. (4) −2a+5, (5) 5a² −2a+1, (6) 3+a

A relation in which there is only one possible pairing of each x and each y is called a function. It should be noted that while the reverse is true, the same y can be paired with different x. Vertical and horizontal lines are the only types of linear equations that are not functions.

Lets evaluate the function a. f(-3)

4. f(-3) = 2(-3)-5

          = −11

5. f(-3) = −5(-3)²+2(-3)−1

          = −52

6.  f(-3) = -3-1-(-3)+1

            = 0

Lets evaluate the function b. f(2)

4. f(2) = 2(2)-5

          = −1

5. f(2) = −5(2)²+2(2)−1

          = −17

6.  f(2) = -3-1-(2)+1

            = −5

Lets evaluate the function c. f(-a)

4. f(-a) = 2(-a)-5

          = −2a−5

5. f(-a) = −5(-a)²+2(-a)−1

          = −5a² −2a−1

6.  f(-a) = -3-1-(-a)+1

            = −3+a

Lets evaluate the function d. -f(a)

4. -f(a) = -(2(a)-5)

          = −2a+5

5. -f(a) = -(−5(a)²+2(a)−1)

          = 5a² −2a+1

6.  -f(a) = -(-3-1-(a)+1)

            = 3+a

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To determine the function that models the data, we need to analyze the relationship between the cost per photo and the number of photos a customer buys. From the given information, we can observe that the cost per photo varies inversely with the number of photos. This implies that as the number of photos increases, the cost per photo decreases, and vice versa.

To model this relationship, we can use the inverse variation equation, which can be expressed as:

y = k/x

Here, y represents the cost per photo, x represents the number of photos, and k is the constant of variation.

Let's examine the data given in the table to find the value of k:

Number of Photos (x) Cost per Photo (y)

10 10

25 4

50 2

100 1

We can see that as the number of photos increases, the cost per photo decreases. We can use any pair of values from the table to solve for k. Let's choose the pair (50, 2):

2 = k/50

Solving for k:

k = 2 * 50 = 100

Now that we have the value of k, we can write the function that models the data:

y = 100/x

Therefore, the function that models the data is y = 100/x, where y represents the cost per photo and x represents the number of photos a customer buys.

Answer:

D - 41.6pi ft.

Step-by-step explanation:

the 16.1 is the radius of the inner circle, add the 4.7 to that and you get 20.8 multiply that by 2 to get the diameter and you get 41. The formula to get circumference is pi x diameter, so D

The sample space is (a) watermelon, watermelon, lemon-lime, lemon-lime, lemon-lime, grape gumballs, grape gumballs, grape gumballs, grape gumballs, grape gumballs

From the question, we have the following parameters that can be used in our computation:

2 watermelon, 3 lemon-lime, and 5 grape gumballs

Rewrite the items according to their frequencies

So, we have the following representation

watermelon, watermelon, lemon-lime, lemon-lime, lemon-lime, grape gumballs, grape gumballs, grape gumballs, grape gumballs, grape gumballs

The above represents the sample space

Hence, the sample space is (a)

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Responder:

x = 4.48

y = 4.86

z = -1.04

Explicación paso a paso:

Dadas las ecuaciones simultáneas;

6X + 8Y - 4Z = 5 .... 1

1X + 2Y + 5Z = 9 ... 2

9X - 3Y + 3Z = 6 .... 3

Primero necesitamos reducir el sistema de ecuaciones

Ecuación 1 * 1

Ecuación 2 * 4

__________

6X + 8Y - 4Z = 5

4X + 8Y + 20Z = 36

Restar

6x-4x - 4z-20z = 5-36

2x-24z = 31 ... 4

también

ecuación 2 * 3

ecuación 3 * 2

_______________

1X + 2Y + 5Z = 9 * 3

9X - 3Y + 3Z = 6 * 2

_______________

3x + 6y + 15z = 27

18x - 6 años + 6z = 12

Agregar

3x + 18x + 15z + 6z = 27 + 12

21x + 21z = 39 ...... 5

Resolviendo 4 y 5

2x-24z = 31 ... 4 * 21

21x + 21z = 39 ...... 5 * 2

____________

42x - 504z = 651

42x + 42z = 78

Sustraer

-504x - 42z = 651-78

-546z = 573

z = -573/546

z = -1.05

Sustituye z = -1.05 en 4

De 4;

2x-21z = 31

2x - 21 (-1,05) = 31

2x + 22,04 = 31

2x = 31-22.04

2x = 8,96

x = 8,96 / 2

x = 4,48

Obtener y

De 2:

X + 2Y + 5Z = 9

4.48 + 2y + 5 (-1.04) = 9

2 años + 4,48-5,2 = 9

2y = 9 + 0,72

2 años = 9,72

y = 9,72 / 2

y = 4,86

Por tanto, la solución (x, y, z) es (4.48, 4.86, -1.04)

Hence, we know that area of rectangle is equals to the product of length and breadth.

Therefore, the area of rectangle is,

\( = 4 \times( {x}^{2} +3x + 2 ) \\ = 4({x}^{2} + 3x + 2) \\ = \green{ \boxed{4 {x}^{2} + 12x + 8}}\)

Therefore, the answer is 4x² + 12x + 8.

Answer:

$659

Step-by-step explanation:

This a basic algebra question with set up.

The total cost is 19,111 with 29 people going and s being each person's cost.

19,111 = 29s

\(\frac{19111} {29}\)\(=s\\\)

so

$659

Answer:

S=$659

Step-by-step explanation:

You have to divide 29 by 29 which give you 1 so cross that out, then do 19111 divided by 29 which is 659, so s=659.

Hope this helps:)

Answer:

X greater than six is the correct answer

Answer:

$4.40

Step-by-step explanation:

you do 24=5x-2 so first you do 24-2 and u get 22 so then divide that by 5 since they each pay 5. then u end up with 4 dollars and 40 cents

Answer:

  C  (x, y) -> (x + 7, y + 8); scale factor = 3

Step-by-step explanation:

The translation can be found by subtracting the coordinates of the original center from those of the center of the image. The scale factor will be the ratio of the radii.

__

We note the center coordinates are I(-3, -6) and I'(4, 2). Then the translation is ...

  I' -I = (4, 2) -(-3, -6) = (4 +3, 2 +6) = (7, 8)

  (x, y) ⇒ (x +7, y +8)

The radius of each circle can be found by counting the grid squares from the center to the circle. It works best to do this along a horizontal or vertical line.

The radius of circle I is 2; the radius of circle I' is 6. The scale factor is ...

  scale factor = 6/2 = 3

Answer:

a. 2, because it is the coefficient of the highest degree.

b. 4

c. 5 terms.

d. -10

e. 3, as it is the coefficient of x^2.

If you need more help don't hesitate to ask :)

Answer:

68276/125

Step-by-step explanation:

500+40+6+1/5+1/125

546+26/125

68276/125

Answer:

546.208

Step-by-step explanation:

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{\color{#c92786}{5}} \cdot {\color{#c92786}{100}}+4 \cdot 10+6 \cdot 1+2x(\frac{1}{10})+8x(\frac{1}{1000})

5⋅100+4⋅10+6⋅1+2x(101​)+8x(10001​)

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500+{\color{#c92786}{4}} \cdot {\color{#c92786}{10}}+6 \cdot 1+2x(\frac{1}{10})+8x(\frac{1}{1000})

500+4⋅10+6⋅1+2x(101​)+8x(10001​)

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500+40+{\color{#c92786}{6}} \cdot {\color{#c92786}{1}}+2x(\frac{1}{10})+8x(\frac{1}{1000})

500+40+6⋅1+2x(101​)+8x(10001​)

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500+40+6+{\color{#c92786}{2}}x \cdot {\color{#c92786}{\frac{1}{10}}}+8x(\frac{1}{1000})

500+40+6+2x⋅101​+8x(10001​)

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500+40+6+\frac{1}{5}x+8x(\frac{1}{1000})

500+40+6+51​x+8x(10001​)

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500+40+6+\frac{1x}{5}+8x(\frac{1}{1000})

500+40+6+51x​+8x(10001​)

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500+40+6+\frac{x}{5}+{\color{#c92786}{8}}x \cdot {\color{#c92786}{\frac{1}{1000}}}

500+40+6+5x​+8x⋅10001​

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500+40+6+\frac{x}{5}+\frac{1}{125}x

500+40+6+5x​+1251​x

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500+40+6+\frac{x}{5}+\frac{1x}{125}

500+40+6+5x​+1251x​

9

Multiply by 1

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500+40+6+\frac{x}{5}+\frac{1x}{125}

500+40+6+5x​+1251x​

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{\color{#c92786}{500}}+{\color{#c92786}{40}}+{\color{#c92786}{6}}+\frac{x}{5}+\frac{x}{125}

500+40+6+5x​+125x​

5

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{\color{#c92786}{546}}+\frac{x}{5}+\frac{x}{125}

546+5x​+125x​

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Solution

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Answer:

no because these "imaginary numbers" can have the same value as 2 real numbers that add up to a real number. Example:7 3+4=7    x+y (Algebra so not real numbers)= infinite possibilities

Step-by-step explanation:

Answer:

68.27%

Step-by-step explanation:

By the Empirical Rule, the amount of data that is 1σ below and above the mean μ accounts for 68.27% of the data.

Step-by-step explanation:

4y(3x+2)–2(3x+2)

(4y–2)(3x+2)

2(2y–1)(3x+2)

The Factor of given equation  \(4y(3x+2)-2(3x+2)\)   is    \((2y-1) (6x+4)\) .

Factor in an equation is something which is multiplied by something else or we can say that numbers or the variables that are multiplied together.

We have,

 \(4y(3x+2)-2(3x+2)\)  

So, first simplify the expression,

      \(12xy+8y -6x-4\)

⇒    \(12xy-6x+8y-4\)

⇒     \(6x(2y-1)+4(2y-1)\)

From we can see that \((2y-1)\) is common in both terms,

Therefore,

⇒    \((2y-1) (6x+4)\)

So,    \((2y-1) (6x+4)\)  is the factor of the given expression.

Hence, we can say that the Factor of given equation \(4y(3x+2)-2(3x+2)\)   is    \((2y-1) (6x+4)\) .

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Answer:

The inverse is,

\(f^{-1}(x) = \sqrt[5]{x} - 3\)

Step-by-step explanation:

f(x) = (x+3)^5

Finding the inverse,

\(f(x) = (x+3)^5\\or,\\y = (x+3)^5\)

We replace x with y and vice versa

so,

\(x = (y+3)^5\)

Solving for y,

the the 5th root,

\(\sqrt[5]{x} = y + 3\\y = \sqrt[5]{x} - 3\)

hence the inverse function is,

\(f^{-1}(x) = \sqrt[5]{x} - 3\)

Step-by-step explanation:

f(x)=(x+3)×5

Y=5X+15

now

interxchanging x and y we get,

x=5y+15

5y=x-15

y=x-15/5

therefor f~1(x)=x-15/5

Answer: D

Step-by-step explanation:

1 in.=1488 milesx5=7440 miles.

Area= pi x r^2

=pi x (7440)^2 or option D.

Hope this helps :)

Answer:

\(R=9\)

\(s=3.0972\)

Step-by-step explanation:

Range (\(R\)) is the highest value minus the lowest value.

The highest value in the data is 10, and the lowest value in the data is 1.

\(R=10-1=9\)

To find the standard deviation of a sample (\(s\)) without a calculator, use the sample variance formula for samples: \(s^{2} =\frac{n(\Sigma{X^2})-(\Sigma{X})^2}{n(n-1)}\), where \({X}\) represents each individual value and \(n\) represents the sample size.

1. Find the sum of the values.

\(\Sigma{X}=1+7+2+1+4+7+3+5+10+9+1+4+3+5+8+1+1+8+5+5+6+10+10+7=123\)

2. Square each value and find the sum.

\(\Sigma{X}^{2}=1^{2}+7^{2}+2^{2}+1^{2}+4^{2}+7^{2}+3^{2}+5^{2}+10^{2}+9^{2}+1^{2}+4^{2}+3^{2}+5^{2}+8^{2}+1^{2}+1^{2}+8^{2}+5^{2}+5^{2}+6^{2}+10^{2}+10^{2}+7^{2}=851\)

3. Substitute it into the sample variance formula.

\(s^{2} =\frac{n(\Sigma{X^2})-(\Sigma{X})^2}{n(n-1)}=\frac{24(851)-(123)^2}{24(24-1)}=\frac{20424-15129}{24(23)}=\frac{5295}{552}=\frac{1765}{184}\)

4. Take the square root of the sample variance to get the sample standard deviation.

\(\sqrt{s^2}=\sqrt{\frac{1765}{184} }=3.097158586\)

5. Round the sample standard deviation to four decimal places.

\(s=3.0972\)

a) To find the number of students getting more than 45 marks, we count the students whose marks are greater than 45 in the given list.

In the given list, the students with marks greater than 45 are: 50, 49, 47, 48, 49, 48, 47, 49, 48, 50, 47, 49, 48, 46, 47, 48, 47.

Counting these numbers, we find that there are 17 students who obtained more than 45 marks.

b) To find the number of students getting less than 45 marks, we count the students whose marks are less than 45 in the given list.

In the given list, the students with marks less than 45 are: 44, 44, 42, 41, 41, 42, 41, 44, 44, 42, 45, 45, 45, 44, 45.

Counting these numbers, we find that there are 15 students who obtained less than 45 marks.

c) To find the maximum number of students getting the same marks, we look for the mark that appears most frequently in the given list.

In the given list, the marks obtained by the students are: 44, 50, 44, 49, 42, 47, 45, 42, 44, 48, 49, 48, 47, 49, 47, 41, 45, 48, 41, 48, 41, 42, 47, 49, 49, 48, 50, 47, 49, 48, 46, 44, 45, 45, 46, 44, 42, 47, 48, 45.

Counting the frequency of each mark, we find that the marks 47 and 48 appear most frequently, with a count of 6 each. Therefore, the maximum number of students getting the same marks is 6.

d) To find the average marks obtained by the students in the class, we sum up all the marks and divide by the total number of students.

Total marks = 44 + 50 + 44 + 49 + 42 + 47 + 45 + 42 + 44 + 48 + 49 + 48 + 47 + 49 + 47 + 41 + 45 + 48 + 41 + 48 + 41 + 42 + 47 + 49 + 49 + 48 + 50 + 47 + 49 + 48 + 46 + 44 + 45 + 45 + 46 + 44 + 42 + 47 + 48 + 45

           = 1912

Total number of students = 40

Average marks = Total marks / Total number of students

                 = 1912 / 40

                 = 47.8

Therefore, the average marks obtained by the students in the class is 47.8.

e) To find the number of students getting more than the average marks, we count the students whose marks are greater than 47.8.

In the given list, the students with marks greater than 47.8 are: 50, 50, 49, 48, 49, 48, 49, 48, 50, 49, 49, 48, 48, 50, 49, 49, 48, 48, 47, 49, 49, 48, 50, 47,

49, 49, 48, 50, 47, 49, 49, 48, 48, 49, 48, 47, 48, 49, 49, 48, 50, 49.

Counting these numbers, we find that there are 40 students who obtained more than the average marks.

If an equivalence relation is a class, then its equivalence classes may be proper classes. This means that when an equivalence relation is considered as a single entity (a class), the individual equivalence classes within that relation may be so large that they cannot be considered as sets, but rather as proper classes.

1. An equivalence relation is a relation that satisfies three properties: reflexivity, symmetry, and transitivity. It is denoted by the symbol "~".

2. In set theory, classes can be divided into two categories: sets and proper classes. Sets are collections of elements, while proper classes are collections that are too large to be considered sets.

3. If an equivalence relation is considered as a class, it means that the relation as a whole is treated as a single entity.

4. The equivalence classes of an equivalence relation are the subsets of the underlying set that contain elements related to each other by the relation.

5. The statement "If an equivalence relation is a class, then its equivalence classes may be proper classes" implies that when the equivalence relation is treated as a class, the individual equivalence classes within that relation may be so large that they cannot be considered as sets.

6. This occurs when the equivalence relation partitions the underlying set into subsets that are proper classes, rather than sets.

7. The reason behind this is that proper classes can be too large to be members of a set, as they would lead to logical paradoxes.

8. Therefore, if the equivalence relation itself is considered as a class, it is possible for its equivalence classes to be proper classes.

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It takes time of near about 35 year.

Given that,

Storage capability of computers has been doubling every 5 years

The date of 1st computer made = 1980

computer could store 5 megabytes,

Now,

We can use the following calculation to determine how long it took computers to store 100 megabytes:

Therefore,

Log₂ (final amount / beginning amount)

=  Log₂ (100 / 0.5)

= log₂ (200)

= 7.64 is the number of doublings.

If each doubling takes five years, then it would take computers just over seven doublings or almost 35 years to store 100 megabytes.

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Answer:

P(Red) = 1/3

Step-by-step explanation:

Total number of marbles = 15

5 red

6 green

4 blue

Answer:

The decision rule will be as follows: Reject the null hypothesis if test-statistic < -2.5 or fail to reject the

Now, Our test-statistic was -1.96, which is greater than -2.5. Thus, we will fail to reject the null hypothesis.

Step-by-step explanation:

We are given;

Population mean; μ = 432

Sample mean; x¯ = 426

Sample size; n = 24

Variance; Var = 225

Now, standard deviation from variance is; σ = √Var

Thus, σ = √225

σ = 15

Hypothesis is defined as;

Null hypothesis; H0: μ ≥ 432

Alternative hypothesis: Ha: μ < 432

Formula for the test statistic since sample size is less than 30 is;

t = (x¯ - μ)/(σ/√n)

Plugging in the relevant values;

t = (426 - 432)/(15/√24)

t = -1.96

Now,we are given significance level of 0.01; our DF = n - 1 = 24 - 1 = 23.

From t-table attached, we can see that at DF of 23 and significance value of 0.01, the critical value is 2.5

However, this is a left tailed test as the rejection region is to the left.

Thus, the critical value of a left tailed test is negative. Thus, in this case our critical value will be -2.5.

The decision rule will be as follows: Reject the null hypothesis if test-statistic < -2.5 or fail to reject the

Now, Our test-statistic was -1.96, which is greater than -2.5. Thus, we will fail to reject the null hypothesis.