If y=2x-1 is changed to y = x+ 2,
how would the graph of the new
function compare with the first one in terms of steepness and shift?

Answer:

4

Step-by-step explanation:

3s=14-2

3s=12

3s/3=12/3

s=4

Answer:

3(x + 4)(x - 5) = 3x² - 3x - 60.

Step-by-step explanation:

3(x + 4)(x - 5)

Combine x + 4 and x - 5 using FOIL, or first, outer, inner, last.

F: x * x = x²

O: x * -5 = -5x

I: x * 4 = 4x

L: 4 * -5 = -20

3(x² - 5x + 4x - 20)
Simplify inside the parenthesis.

3(x² - x - 20)

Distribute the 3 into the equation.

3x² - 3x - 60

Answer:  -∞ < x < ∞

Step-by-step explanation:

      The domain is all the x-values, or inputs, of a function. The range is the y-values, or outputs, of a function.

       In this case, we are looking for the x-values of the function. Since the function will keep going "out forever" in both directions, the answer should be the fourth option;

             - ∞ < x < ∞

The first step that we must take to solving this problem is to fully understand what the problem statement is asking from us and what is given us to solve the problem.  Looking at the problem statement, we can see that they are asking us to determine what the domain of the function is in the graph that was provided.  However, first of all, let us define what domain is.

Looking back at our problem, we can see that this is similar to the example that was provided in the definition.  We can see that the parabola reaches out to both positive and negative infinity in the x-direction but at a slope. Although we can only see it reach -2 and 6 we know that the parabola continues going on even after that.  

Therefore, looking at the options that were provided option D, -∞ < x < ∞ would be the best fit as it showcases that x reaches from negative infinity to positive infinity.

You could vote on it? There are as many ways as you can think of. It’s not set in stone.

[tex](\stackrel{x_1}{-3}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{4}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{4}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{(-3)}}} \implies \cfrac{0}{6 +3}\implies 0[/tex]

hmmm a line with a slope of 0 is a horizontal one, and it has no slope-intercept form.

Check the picture below.

Answer:

5/9

Step-by-step explanation:

20÷36

20/36

divide using 2

10/18

divide using 2

5/9

Answer:

3x^2

Step-by-step explanation:

Area = (length) x (width)

Area = (1x)*(3x)

Area = 3x^2

Answer:

10 cards

Step-by-step explanation:

4% of 250 cards

(4/100) x 250 cards

10 cards ✔️

Answer:

2x^3 y^18

Step-by-step explanation:

The power of a power (properties of exponents) was used in this problem. Hope this helps! :)

Answer:

[tex]2x^3 y^{18}[/tex]

Step-by-step explanation:

[tex]~~~\left( 8 x^9 y^{54} \right)^{\tfrac 13}\\\\=\left(2^3 x^9 y^{54} \right)^{\tfrac 13} \\\\=\left( 2^3 \right)^{\tfrac 13}\left( x^9 \right)^{\tfrac 13}\left( y^{54} \right)^{\tfrac 13}\\\\=2^{\tfrac 33} \cdot x^{\tfrac 93} \cdot y^{\tfrac{54}3}\\\\=2x^3 y^{18}[/tex]

Answer:

Step-by-step explanation:

[tex]\sf \left(2x+\cfrac{1}{2}\right)+\left(8x-2\cfrac{1}{2}\right)[/tex]

We'll need to convert the mixed number 2 1/2 to improper fraction.

[tex]\sf \left(2x+\cfrac{1}{2}\right)+\left(8x-\cfrac{5}{2}\right)[/tex]

Simplify...

[tex]\sf 2x+\cfrac{1}{2}+8x-\cfrac{5}{2}[/tex]

Add similar elements.

[tex]\sf 10x+\cfrac{1}{2}-\cfrac{5}{2}[/tex]

[tex]\sf 10x+\cfrac{-4}{2}[/tex]

Divide numbers..

[tex]\sf 10x-2[/tex]

_______________________

The time it would take for the bacteria to reach 1119 is 5.83 hours

Given the function:

[tex]Pt= P_0e^k^t[/tex]

To find the exponential growth;

80 = [tex]50e^k^*^3[/tex]

[tex]k= 1.6/3\\k= 0.533[/tex]

If there is a constant growth, then to get to 1119:

1119= [tex]50e^0^.^5^3^3^*t[/tex]

3.8 = 0.533*t

t= 5.83 hours

Therefore, it would take 5.83 hours for the bacteria to increase to 1119.

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The complete question is

"Segment MP is a diameter of circle O. Circle O is shown. Line segment M P is a diameter. Line segment N O is a radius. Angle N O P is 150 degrees.

Which equation can be used to find mArc M N?

mArc M N + 150 = 180, mArc M N + 150 = 360, mArc M N – 150 = 180, mArc M N – 150 = 360"

The Equation m Arc(MN) + 150  = 180 is used to find mArc MN. Thus, the correct option is A.

A system of equations is two or more equations that can be solved to get a unique solution. the power of the equation must be in one degree.

It is Given:

Segment MP is the diameter of circle O. Circle O is shown.

Line segment M P is a diameter.

Line segment N O is a radius.

∠NOP = 150°

We need to Find the Equation for m Arc MN

We already know the Diameter subtends 180° as it is the half of Circle 360°

∠NOP = 150°  

∴ m(Arc NP) = 150°

The measure of an arc is the measure of its central angle

Therefore,

m Arc Diameter = 180°

m Arc(MN) + m Arc(NP)  = 180

m Arc(MN) + 150  = 180

Therefore, the Equation m Arc(MN) + 150  = 180 is used to find mArc MN.

Thus, the correct option is A.

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Answer:

y = 5sin(6π/7) + 3

Answer:

74

Step-by-step explanation:

(8x10) - 6 =

80 - 6 = 74

653/8

=81 R 5

Step-by-step explanation:

when you divide 653 by 8 you get 82 reminder 5

to prove try multiplying 82 by 8 you will get 5 as the remainder

Answer:

4^3 = 4x4x4

Step-by-step explanation:

its just a shortened version of 4x4 3 times

Answer:

AC = BC = 5

AB = 5√2

∠A = ∠B = 45

∠A = 90

Step-by-step explanation:

AC = 5 ( Reason : 5 squares are present in between A and C )

Similarly,

BC = 5

By Pythagoras theorem,

(AB)² = (AC)² + (BC)²

        = 5² + 5²

        = 2 * 5²

(AB)² = 2 * 5²

AB = 5√2

Since, sides AC and BC are equal,

∠A = ∠B = 45

Since, AC is perpendicular to BC,

∠A = 90

Answer:

-9

Step-by-step explanation:

-6×3/2

-6×3=-18

-18/2

-9

The zeroes of the given polynomial is 1 and -6.

A polynomial is an expression consisting of indeterminates and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.

Here, given equation;

          f(x) = (x - 1)(x + 6)

for finding the zeros of the equation we need to put.....f(x) = 0

           0 = (x - 1)(x + 6)

     x - 1 = 0      or       x + 6 = 0

     x = 1          or    x = -6

Thus, The zeroes of the given polynomial is 1 and -6.

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Answer:

The Graph is included Bellow...

Step-by-step explanation:

This function will start at (0,2) and proceed to the right (based on its family function of √x) and since it has been shifted up 2 from f(x) then we will move two units up.

The domain can be considered the range from right to left that the function is restricted in. In this case, it starts at -2 (Which the function doesn't reach) and will stop at 7.

So the Graph you are looking for will start at (0, 2) and proceed to the right until (7, 4.6) ish. I have included a table to help your graph it!

Please Like if this helps!

Answer:

-23

Step-by-step explanation:

Answer:

5

Step-by-step explanation:

to find the mean you have to add up all the numbers so

1+8=9

9+3=12

12+2=14

14+8=22

22+8= 30

and then you divide by the amount of numbers which is 30 / 6 = 5

Hope This Helped

[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\textbf{QUESTION READS}[/tex]

[tex]\large\textsf{Find the mean from the following data:1, 8, 3, 2, 8, 8}}[/tex]

[tex]\huge\textbf{A LITTLE BREAKDOWN OF MEAN}[/tex]

[tex]\large\text{To find the \underline{\underline{mean}} you have to ADD all of the numbers \& DIVIDE}\\\large\text{by the total numbers you have in your data plot..}[/tex]

[tex]\large\text{Here's the formula: }\large\boxed{\rm{\dfrac{the\ sum\ of\ all\ terms}{number\ of\ terms\ you\ have} = your\ \bf mean}}[/tex]

[tex]\huge\textbf{ANSWERING THE QUESTION}[/tex]

[tex]\large\text{Your equation: }\large\boxed{\rm{\dfrac{1 + 8 + 3 + 2 + 8 + 8}{6} = your \ \bf mean}}\\\\\large\boxed{\rm{\dfrac{1 + 8 + 3 + 2 + 8 + 8}{6}}}\\\\\large\boxed{\rm{= \dfrac{9 + 3 + 2 + 8 + 8}{6}}}\\\\\large\boxed{\rm{= \dfrac{12 + 2 + 8 + 8}{6}}}\\\\\large\boxed{\rm{= \dfrac{14 + 8 + 8}{6}}}\\\\\large\boxed{= \rm{\dfrac{22 + 8}{6}}}\\\\\large\boxed{= \rm{\dfrac{30}{6}}}\\\\\large\boxed{\rm{= 5}}[/tex]

[tex]\huge\textbf{MET THE ANSWER OF THE}\\\huge\textbf{QUESTION}[/tex]

[tex]\huge\text{Therefore, your answer should be: \boxed{\mathsf{5}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

Answer:

[tex]\boxed{\huge{\sf \dfrac{1}{4} }}[/tex]

Explanation:

Given Total Numbers: 8 numbers

Must be greater than 6, The numbers are 7 and 8 = 2 numbers

Probability:

→ favorable outcomes/total possible outcomes

→ 2/8

→ 1/4

→ 0.25

Answer:

168cmsqured

Step-by-step explanation:

as for me,15 acts as the lower part of the trapezium while 13 acts as the upper part of the trapezium .so since I know the measurements of the both sides I can easily calculate the area use using the formula indicated in the image.hope it helps

The option that is the functions is graphed below is known as option A = Y= (x+4) -2.

The function graphed above is Y=|x+4|-2 because  when you solve for the graphs,  the number in the x-axis is always changed or turned to  the other side and there one can see that it would be -4 would be +4, and +4 and also -4.

Note that

Y = (x+4) -2.

x = (0 + 4) - 2

x = 4-2

x =2

Therefore, plug x into the equation.

Y = (x+4) -2.

= ( 2 + 4) - 2

= (6) -2

= 4

Therefore, The option that is the functions is graphed below is known as option A = Y= (x+4) -2.

See full question in the image attached.

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a) If any values in the data set are greater than the upper boundary, then those values are the upper outliers.

b) If any values in the data set are less than the lower boundary, then those values are lower outliers.

The complete question is attached with an answer.

If a number in distribution is more than 1.5 times the length of the box away from either of the upper and lower quartile of the distribution, then that number is said to be an outlier.  

The lower quartile is  and the upper quartile of distribution is

Then lower boundary is given by [tex]=Q_1-1.5\times IQR[/tex]

The Upper boundary is given by  [tex]=Q_3+1.5\times IQR[/tex]

If any number or value of the distribution is greater than  [tex]=Q_3+1.5\times IQR[/tex]  then that is called Upper Outlier.

Similarly, if any value or number of the distribution is less than [tex]=Q_1-1.5\times IQR[/tex] then that is called Lower Outlier.

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