4.
Solve the system of inequalities graphically. Label the solution set with an S.
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Answer:

D

Step-by-step:

The y-coordinates will change, not the x. Size and shape will also remain the same.

Evaluation:

The answe for the function is -1.

The sample mean for the data is 21.38 and the sample standard deviation for the data is 56.19

To do this, we start by calculating the midpoint of each class using:

Midpoint= (Lower + Upper)/2

Using the above formula, we have:

Age (x)        Frequency (f)

2                   24.8

9.5               422

17                  194

22                 256

29.5             50.7

39.5              35.7

54.5             75.2

70                 54.3

This is calculated using:

[tex]\bar x = \frac{\sum fx}{\sum f}[/tex]

So, we have:

[tex]\bar x = \frac{2*24.8 + 9.5*422 + 17*194 + 22*256 + 29.5*50.7 + 39.5*35.7 + 54.5*75.2+70*54.3}{24.8 + 422 + 194 + 256 + 50.7 + 35.7 + 75.2 + 54.3}[/tex]

Evaluate

[tex]\bar x = \frac{23793.8}{1112.7}[/tex]

[tex]\bar x = 21.38[/tex]

Hence, the sample mean for the data is 21.38

This is calculated using:

[tex]\sigma_x = \sqrt{\frac{\sum f(x- \bar x)^2}{\sum f - 1}}[/tex]

So, we have:

[tex]\sigma_x = \sqrt{\frac{2*(24.8-21.38)^2 + .............+70*(54.3-21.38)^2}{24.8 + 422 + 194 + 256 + 50.7 + 35.7 + 75.2 + 54.3 - 1}}[/tex]

Evaluate

[tex]\sigma_x = \sqrt{\frac{3509508.4556}{1111.7}}[/tex]

[tex]\sigma_x = \sqrt{3156.88446128}[/tex]

[tex]\sigma_x = 56.19[/tex]

Hence, the sample standard deviation for the data is 56.19

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Answer: (4,-4)

Step-by-step explanation:

Reflection across the x axis means [tex](x,y) \longrightarrow (x,-y)[/tex], so [tex](3,7) \longrightarrow (3,-7)[/tex].

Reflection across the y axis means [tex](x,y) \longrightarrow (-x,y)[/tex], so [tex](3,-7) \longrightarrow (-3,-7)[/tex].

Then, [tex](-3,-7) \longrightarrow R'\boxed{(4,-4)}[/tex]

Answer:

4hours

Step-by-step explanation:

1hour =85+85

=170km/h

time= distance÷

speed=170km/h

680÷170=time

=4h

Answer:

When -3v = -18 is solved, the result is:

21v

The result of this mathematical sentence is v=6

First degree equation is a mathematical sentence - having at least one unknown. Its representation is given by ax + b = 0. To calculate an equation, as it is in the expression in the statement, simply: isolate x and divide the second to the first member.

*Note: Equal signs, the result will be equated as positive.

-3v = -18

v = -18/-3

v = 6

Therefore, the correct value of this equation is v = 6.

Explanation:

Given f(x) : (2, -3)

f(x) + 2 then graph translates up by 2 units up =  [tex]\boxed{\sf (2, -1)}[/tex]

f(x) - 3 then graph translates down 3 units down = [tex]\boxed{\sf (2, -6)}[/tex]

f(x + 5) then graph translates left 5 units = [tex]\boxed{\sf (-3, -3)}[/tex]

-f(x) then graph reflects over x axis = [tex]\sf \boxed{\sf (2, 3)}[/tex]

f(-x)  then graph reflects over y axis = [tex]\sf \boxed{\sf (-2,-3)}[/tex]

f(2x) then graph has horizontal compression = (2/2, -3) = [tex]\boxed{\sf (1, -3)}[/tex]

2f(x) then graph has vertical compression = (2, (-3)2) = [tex]\boxed{\sf (2, -6)}[/tex]

-f(x - 4) then graph reflects over x axis, moves 4 units to right = [tex]\sf \boxed{\sf (6, 3)}[/tex]

Parent function: y = x²

Graph function: f(x) = (x + 8)² - 4

After Identification:

D. The graph has a translation of 8 units left and 4 units down.

Answer:

Translations

For [tex]a > 0[/tex]

[tex]f(x+a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units left}[/tex]

[tex]f(x-a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units right}[/tex]

[tex]f(x)+a \implies f(x) \: \textsf{translated}\:a\:\textsf{units up}[/tex]

[tex]f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}[/tex]

[tex]y=a\:f(x) \implies f(x) \: \textsf{stretched parallel to the y-axis (vertically) by a factor of}\:a[/tex]

[tex]y=f(ax) \implies f(x) \: \textsf{stretched parallel to the x-axis (horizontally) by a factor of} \: \dfrac{1}{a}[/tex]

[tex]y=-f(x) \implies f(x) \: \textsf{reflected in the} \: x \textsf{-axis}[/tex]

[tex]y=f(-x) \implies f(x) \: \textsf{reflected in the} \: y \textsf{-axis}[/tex]

Given:  [tex]f(x)=(2,-3)[/tex]

[tex]f(x)+2 \implies (x, y+2)= (2,-3+2)=(2,-1)[/tex]

[tex]f(x)-3 \implies (x,y-3)=(2,-3-3)=(2,-6)[/tex]

[tex]f(x+5)\implies (x-5,y)=(2-5,-3)=(-3,-3)[/tex]

[tex]-f(x) \implies (x,-y)=(2,-(-3))=(2,3)[/tex]

[tex]f(-x) \implies (-x,y)=(-(2),-3)=(-2,-3)[/tex]

[tex]f(2x) \implies \left(\dfrac{x}{2},y\right)=\left(\dfrac{2}{2},-3\right)=(1,-3)[/tex]

[tex]2f(x) \implies (x,2y)=(2,2 \cdot -3)=(2,-6)[/tex]

[tex]-f(x-4) \implies (x+4,-y)=(2+4,-(-3))=(6,3)[/tex]

[tex]\begin{array}{| c | c | c | c | c | c | c | c |}\cline{1-8} & & & & & & &\\f(x)+2 & f(x)-3 & f(x+5) & -f(x) & f(-x) & f(2x) & 2f(x) & -f(x-4)\\& & & & & & &\\\cline{1-8} & & & & & & &\\(2,-1) & (2,-6) & (-3,-3) & (2,3) & (-2,-3) & (1,-3) & (2,-6) & (6,3)\\& & & & & & &\\\cline{1-8} \end{array}[/tex]

Parent function:  [tex]y=x^2[/tex]

Given function:  [tex]f(x)=(x+8)^2-4[/tex]

[tex]f(x+8) \implies f(x) \: \textsf{translated}\:8\:\textsf{units left}[/tex]

[tex]f(x)-4 \implies f(x) \: \textsf{translated}\:4\:\textsf{units down}[/tex]

Therefore, a translation 8 units to the left and 4 units down.

Answer:

Use prime factorizations to find the GCF of 18 and 33.

Prime factorization of 18: 2 × 3 × 3

Prime factorization of 33: 3 x 11

The GCF of 18 and 33 is

Step-by-step explanation:

Use prime factorizations to find the GCF of 18 and 33.

Prime factorization of 18: 2 × 3 × 3

Prime factorization of 33: 3 x 11

The GCF of 18 and 33 is

Answer:

3 XD

Step-by-step explanation:

Answer:

Solution = (2,1)

Step-by-step explanation:

Make them equal each other and solve for x.

2x - 3 = 1

2x = 4

x = 2

2 * 2 - 3 = 1

Solution = (2,1)

[tex]3(x - 8) - 13x(x - 8) + 3 = 0 \\ 3x - 24 - 13 {x}^{2} + 104x + 3 = 0 \\ - 13 {x}^{2} + 107x - 21 = 0 \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a } \\ a = - 13 \\ b = 107 \\ c = - 21 \\ x = \frac{ - 107 + - \sqrt{ {107}^{2} - 4(13)( - 21)} }{2( - 13)} \\ x1 = \frac{107 - \sqrt{10357} }{26} \\ x2 = \frac{107 + \sqrt{10357} }{26} [/tex]

Hope it helps

Please give brainliest

find D which is the discriminant and with D < 0.

therefore the equation has an imaginary root and not a real root.

Answer:

Step-by-step explanation:

Comment

The way the diagram looks when it is drawn is that you are going to need the sine function to solve for the side opposite which is CD.

Formula

Sin(A) = Sin(68 = opposite / hypotenuse     Substitute values.

Solution

sin(68) = altitude (side opposite) / 20       multiply both sides by 20

20*sin(68) = 20* altitude / 20                     substitute for sin(68)

20* 0.9272 = altitude

altitude = 18.5

Answer: 18.5 to the nearest tenth

Answer:

D is the answer hope

Step-by-step explanation:

hope this helped

Answer:

[tex]\huge\boxed{\sf R. 16.75}[/tex]

Step-by-step explanation:

*Assuming commas are used for decimals.

2.6 kg potatoes = R 13.20

Divide 2.6 to both sides

1 kg potatoes = R 13.20/2.6

1 kg potatoes = R 5.1

Multiply 3.3 to both sides

3.3 kg potatoes = R 5.1 × 3.3

3.3 kg potatoes = R 16.75

[tex]\rule[225]{225}{2}[/tex]

Answer:

its C ^^ makes more sense :D

Step-by-step explanation:

hope this helps

Answer:

the leftover piece of ribbon is 5 3/4 feet long.

Step-by-step explanation:

if she has 7 1/2 feet of ribbon, and she removes (by cutting it off) 1 3/4 feet of ribbon.

So, if we set up this scenario like a math problem, we can write it as this:

7 1/2 feet of ribbon - 1 3/4 feet of ribbon = remaining ribbon

[ - = take away]

so, we are solving the equation:

7 1/2 - 1 3/4

To solve this, it is easiest to make the fractions have the same denominators

1 / 2 is equal to 2 / 4

However, 2/4 is less than 3/4, so we need to make this fraction a larger number.

To do this, we can take away 1 foot from the 7, and write it as if it were fourths

1 [foot] = 4/4 quarter foot

So, we can add 4/4 + 2/4 = 6/4

Now, we can re-write our situation.

6 6/4  -   1 3/4 = left over ribbon

   6      6/4

-   1       3/4

__________

   5      3/4

So, the leftover piece of ribbon is 5 3/4 feet long.

Hope this helps!

Profit percentage for weaver after selling mat is equals to 140%.

" Percentage is defined as the hundredth part of the whole given quantity. It is represented by symbol %."

Formula used

Profit% = [tex](\frac{Profit }{cost price} )[/tex]×   100

According to the question,

Given,

Cost price of bundle of grass = $50.00

Number of mats made = 8

Selling price of each mat = $15.00

Therefore,

Selling price of 8 mats = 15 × 8

                                     = $120

Selling price > Cost price

Therefore,

Profit = Selling price - Cost price

         = 120 - 50

         = $70

Substitute the value in the formula to get profit percentage we get,

Profit% = [tex]\frac{70}{50}[/tex] ×100

            = 140%

Hence, profit percentage for weaver after selling mat is equals to 140%.

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Answer:

79.8

Step-by-step explanation:

Answer:

62.

Step-by-step explanation:

because I subtract 70 to 8 equal to 62

Answer:

I think

Step-by-step explanation:

I think 62 hope you have a good day

Answer:

BB

Hope that helps!

Answer:

x=5.24-9+180n,174.75+180n

Step-by-step explanation:

Answer:

m = -3

Step-by-step explanation:

The formula to find the slope of the line is :

slope = m = [tex]\frac{y_1 - y_2}{x_1-x_2}[/tex]

Given that the two coordinates of the line are :

( -1 , - 7 ) ⇒ ( x₁ , y₁ )

( 1 , -13 )  ⇒ ( x₂ , y₂ )

Let us solve now.

m = ( y₁ - y₂ ) ÷ ( x₁ - x₂ )

m =  ( -7 - ( -13)) ÷ ( -1 - 1 )

m = ( -7 + 13 ) ÷ ( -2 )

m = 6 ÷ -2

m = -3

Answer:

m = -3

Step-by-step explanation:

Given two points:

To Find:

Solution:

Using slope's formulae,

[tex] \boxed{ \rm{m = \cfrac{y_2 -y_1 }{x_2 - x_1} }}[/tex]

According to the question, on the formula:

Substitute them onto the formulae:

[tex] \rm \: m = \cfrac{ - 13 - ( - 7)}{1 - ( - 1)} [/tex]

Simplify using PEMDAS:

[tex] \rm \: m = \cfrac{ - 13 + 7}{1 \ + 1} [/tex]

[tex] \rm \: m = \cfrac{ - \cancel6}{ \cancel2} = \boxed{ - 3}[/tex]

Hence, the slope of the line that passes through the given points in it's simplest form is -3.

Answer:

Step-by-step explanation:

What is 20 divided by 1/2 inches

20 : 1/2 =

20 * 2 =

40

The range of a graph of a function is (-∞, 3) U (3, ∞) option (D) (-∞, 3) U (3, ∞) is correct.

It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

It is given that:

The graph of a hyperbola function is given.

As we know the graph of an inverse of a function is reflections over the line y=x.

We get the same graph as given in the picture.

The range of a graph of a function:

Range: (-∞, 3) U (3, ∞)

Thus, the range of a graph of a function is (-∞, 3) U (3, ∞) option (D) (-∞, 3) U (3, ∞) is correct.

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The table that represents an exponential function of the form y = [tex]b^{x}[/tex] when 0 < b < 1 is Table - 2. See the attached tables.

A function is exponential when its value is a constant that is raised to the power of the argument. This is so especially when the function of the constant is e.

Recall that the exponential function y = [tex]b^{x}[/tex] given that 0 < b < 1. Notice that the table number 2 see to the exponential function that has the following form:

y(x) = (1/3)ˣ

substituting the values of x into the equation, we have:

y(-3) = (1/3) ⁻³ = 27

x = -2; thus

y(-2) =  (1/3) ⁻³ = 9

x = -1; thus

y(-1) =  (1/3) ⁻³ = 3

x = 0; thus

y(0) =  (1/3) ⁻⁰ = 1

x = 1; thus

y(1) =  (1/3) ¹ = 1/3

x = 2; thus

y(2) =  (1/3) ⁻² = 1/9

x = 3; thus

y(3) =  (1/3) ⁻³ = 1/27

Therefore, according to the obtained values, one can summarize that the table that depicts the exponential function y  = bˣ is table 2.

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The update of the completed graph is given in the image attached as it shows the possible combinations of Yellow or Xanadu party favors that might choose will be ( 48, 0).

For this graph, we shall make Xanadu party favors  as X

and yellow favors  as  Y

Therefore note that: X + Y = 48

In the graph of X - Intercept

Y = 0, X = 48

In the graph of Y- intercept,

Y = 48 , X = 0

Therefore, the graphical representation of the combinations is given in the image attached.

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