Which transformations could have taken place? Select two options.

Answer:

30 dollars

Step-by-step explanation:

25/5 = 5 dollars per hour

so 6 hours would be 5 * 6 = 30 dollars

Answer:

It would be B

Step-by-step explanation:

First, we need to find out the unit rate

5/25=1/5

Therefore, that means in one hour the cost would be $5

Since you are going from one hour to 6 hours you need to multiply both numbers by 6

Since 5x6=30, it would be 30$ :)

Answer:

Decreases

Step-by-step explanation:

A negative correlation means the y is big when the x is small. As x gets bigger, y gets smaller.

The description of the correlation in terms of strength (weak or strong) and direction positive or negative) is as explained below.

A correlation coefficient, is expressed as r and it indicates a measure of the direction and strength of a relationship between two variables.

Now, in terms of strength, Correlation strength ranges from -1 to +1. Thus, The closer the correlation is to 0, the weaker it is while the closer it is to ±1, the stronger it is.

Now, in terms of direction;

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The equation of the appropriate line of best fit is y = 1.25x + 2

Start by drawing a line through the points (see attachment)

From the attached graph, we have:

(x, y) = (0,2) and (4,6.5)

Calculate the slope (m) using:

m = (y2 - y1)/(x2 - x1)

This gives

m = (6.5 - 4)/(2 - 0)

Evaluate

m = 1.25

The equation is then calculated as:

y = m(x - x1) + y1

This gives

y = 1.25(x - 0) + 2

Expand

y = 1.25x + 2

Hence, the equation of the appropriate line of best fit is y = 1.25x + 2

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The attached graph represents the function h(x), where h(x) = f(x)g(x)

Consider the functions f(x) = x - 6 and g(x) = x + 6. Which graph shows h(x) = f(x)g(x)?

We have

f(x) = x - 6

g(x) = x + 6

The equation h(x) = f(x)g(x) means

h(x) = f(x) * g(x)

So, we have:

h(x) = (x - 6) * (x + 6)

Expand

h(x) = x² - 36

See attachment for the graph of function h(x)

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Answer:

A on edg. 2020

Step-by-step explanation:

Which graph shows h(x) = f(x)/g(x)

The solution of the equation -x^2 + 2x + 3 = x^2 - 2x + 3 are the x-coordinates of common points on the graphs are x=0 and x=2.

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.

As we can see from the attached graph these graphs intersect at points (0,3) and (2,3).

The system of equation are;

[tex]y = x^2 + 2x + 3 \\y = x^2 - 2x + 3[/tex]

So,

[tex]-x^2 + 2x + 3 = x^2 - 2x + 3 \\\\2x^2 - 4x = 0\\\\2x ( x - 2) = 0[/tex]

Then the solutions of the equation are x=0 and x=2.

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Answer:

0 and 2

Step-by-step explanation:

b on edge

Answer:

[tex]1\frac{1}{5}[/tex]

Step-by-step explanation:

3 = 3/1

(multiply top and bottom separately for fractions)

[tex]\frac{3}{1}[/tex] × [tex]\frac{2}{5}[/tex] = [tex]\frac{6}{5}[/tex]

[tex]\frac{6}{5} = \frac{5}{5}+\frac{1}{5}= 1\frac{1}{5}[/tex]

Answer:

Step-by-step explanation:

multiplying three (3) and two(2) and express the product of three and two over five.

five can enter in six three times , remainder 1

The fraction in lowest terms is one whole number , one over five.

Answer:

4 hours

Step-by-step explanation:

First we need to subtract the one pay fee (the helmet rental fee, $15)

So 51 - 15 = 36

Then we need to divide 36 by the amount of money it costs for each hours to find how many hours he rented the bike

So 39/9 = 4

Answer:

27/4

Step-by-step explanation:

added in the picture

Bearing is a topic that deals with the distance and location of a given object with respect to a reference point. Thus in the given question, the ship is 57.23 km to the yacht.

Bearing is a topic that considers the distance and position of a given object with respect to a reference point. Thus the angles are usually measured with reference to the north pole.

Thus to solve the given question, let the required distance be represented by x.

Applying the cosine rule, we have;

[tex]c^{2}[/tex] = [tex]a^{2}[/tex] + [tex]b^{2}[/tex] - 2ab Cos C

The angle C between the line from ship to tugboat and from tugboat to yacht can be determined as;

C = [tex]10^{o}[/tex] + [tex]40^{o}[/tex]

C = [tex]50^{o}[/tex]

Thus,

[tex]c^{2}[/tex] = [tex]36^{2}[/tex] + [tex]27^{2}[/tex] + 2(36 x 27) Cos [tex]50^{o}[/tex]

  = 1296 + 729 + 1944(0.6429)

  = 2025 + 1249.8

[tex]c^{2}[/tex] = 3274.8

c = [tex]\sqrt{3274.8}[/tex]

  = 57.2259

c = 57.23

Therefore, the distance from the ship to the yacht is 57.23 km.

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Kindly contact a 1-on-1 tutor if more explanations are needed.

Answer:

10.33332

Step-by-step explanation:

-6.77777 is 6.77777 units away from 0. So there's 6.77777+3.55555 which is 3.55555 units away from 0.

Add those two numbers together and it's 10.33332

Answer:

10.33332

Step-by-step explanation:

distance between -6.77777 and 3.55555 is 10.33332

Answer:

The answer is 105.

Step-by-step explanation:

Step-by-step explanation:

A function is continuous at x is

Limit as x approaches c= f(c).

Step 1: Find f(4).

4 is defined for 2x-1 so plug in. 4 for x.

[tex]2(4) - 1 = 7[/tex]

Note that what we just calculated was not only f(4) but also the left sided limit of f(x).

Step 2: Make sure the left sided and right sided limit are equal to 7.

[tex]a + 5x = 7[/tex]

Plug in 4 for x

[tex]a + 5(4) = 7[/tex]

[tex]a = - 13[/tex]

If [tex]x=p+2[/tex] and [tex]y=p-3[/tex], then

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

and

[tex]x-y = (p+2)-(p-3) = 5 \implies x-y-5=0[/tex]

as required.

Answer: B. 4x-2y=2

Step-by-step explanation:

We will work this out one by one. Plug in the x and y values of the solution to the equations and you will see which one is true or false.

A. [tex]2(1)+y(1)=2\\2+1\neq 2[/tex]     B. [tex]4(1)-2(1)=2\\4-2=2[/tex]

C. [tex]2(1)-2(1)=2\\2-2\neq 2[/tex]     D. [tex]x(1)+y(1)=4\\1+1\neq 4[/tex]

With this information, we see that B is the correct answer.

I hope this helped!! Pls mark brainliest!! :))

Answer:

P=3.5

P=2.667

Step-by-step explanation:

p=f/a

P=70/20

P=3.5

then f is increased by 10

f=70+10=80

and a is increased by 10

a=20+10

a=30

therefore; P=f/a

P=80/30

P=2.667

Answer:  [tex]x^2[/tex]  and  [tex]-2[/tex]

Step-by-step explanation:

original :

[tex](10x^3+3x^2)+(-20x-6)[/tex]

factored :

[tex]x^2(10x+3)+-2(10x+3)[/tex]

Answer :

[tex]x^2[/tex]  and  [tex]-2[/tex]

Using the given information, the smallest possible value of x - z is 12

From the question, we are to determine the smallest possible value of x - z

From the given information,

w and x are whole numbers

and

w > 60

x < 50

∴ w = 61, 62, 63, 63, 65, ...

x = 49, 48, 47, 46, 45, ...

Then,

The smallest possible value of x - z is

x - z = 61 - 49

x - z = 12

Hence, the smallest possible value of x - z is 12

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Answer: cosecant

Formula's are:

Secant:

1/cos(x) = sec(x)

Cosecant:

1/sin(x) = cosec(x)

Cotangent:

1/tan(x) = cos(x)/sin(x) = cot(x)

Answer:

8

Step-by-step explanation:

If the only zero for this function is at x = - 4 it looks like this:

f(x)  = (x+4)(x+4)        =    x^2 +8x + 16         so j = 8

Answer:

Step-by-step explanation:

x² +jx +k = 0

If there is only one root ( - 4 ), then (x + 4)² = 0

or

x² + 8x + 16 = 0

j = 8 and k = 16

[tex](a+b)(a-b) = a^2 - ab + ba - b^2 = a^2 - b^2[/tex]

Since the middle terms cancel each other out, what is left is the difference of two squares.

In your example (x+6)(x-6) = x² -36

Answer: The second option, y = 3(1/3)^x

Step-by-step explanation: Input each equation into a graphing calculator (preferably GeoGebra online) and check the answer. I linked the equation, but make sure to check anyway just in case :)

Answer:

[tex]x \approx 2.740646096[/tex]

Step-by-step explanation:

Given:

[tex]x\log_{10}x=1.2[/tex]

Rewrite so that it's equal to zero:

[tex]\implies x\log_{10}x-1.2=0[/tex]

Therefore:

[tex]\implies f(x)=x\log_{10}(x)-1.2[/tex]

Differentiate the function:

Use the product rule to differentiate [tex]x\log_{10}(x)[/tex]:

[tex]\textsf{Let }u=x \implies \dfrac{du}{dx}=1[/tex]

[tex]\textsf{Let }v=\log_{10}(x) \implies \dfrac{dv}{dx}=\dfrac{1}{x \ln 10}[/tex]

[tex]\begin{aligned}\implies \dfrac{dy}{dx} & = u\dfrac{dv}{dx}+v \dfrac{du}{dx}\\& = \dfrac{x}{x \ln 10}+\log_{10}(x)\\& = \dfrac{1}{\ln 10}+\log_{10}(x)\end{aligned}[/tex]

[tex]\implies f'(x)=\dfrac{1}{\ln 10}+\log_{10}(x)[/tex]

[tex]\implies f'(x)=\dfrac{1+\ln 10 \log_{10}(x)}{\ln 10}[/tex]

Newton-Rhapson iteration formula

[tex]x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}[/tex]

Substitute the function and its derivative into the formula, replacing x with [tex]x_n[/tex]:

[tex]\implies x_{n+1}=x_n-\dfrac{x_n\log_{10}(x_n)-1.2}{\dfrac{1+\ln 10 \log_{10}(x)}{\ln 10}}[/tex]

[tex]\implies x_{n+1}=x_n-\dfrac{x_n\ln 10\log_{10}(x_n)-1.2\ln 10}{1+\ln 10 \log_{10}(x_n)}[/tex]

Therefore, the iteration formula is:

[tex]x_{n+1}=x_n-\dfrac{x_n\ln 10\log_{10}(x_n)-1.2\ln 10}{1+\ln 10 \log_{10}(x_n)}[/tex]

Let [tex]x_0=3[/tex]

Substitute this into the formula and carry out 3 iterations:

[tex]\implies x_{1}=3-\dfrac{3\ln 10\log_{10}(3)-1.2\ln 10}{1+\ln 10 \log_{10}(3)}=2.746149035[/tex]

[tex]\implies x_2=2.740648841[/tex]

[tex]\implies x_3= 2.740646096[/tex]

x = the quantity of 3 plus or minus the square root of 29 all over 2. Then the correct option is B.

The quadratic equation is given as ax² + bx + c = 0.

Then the solution is given as

[tex]\rm x = \dfrac{-b\pm \sqrt{b^2 - 4ac}}{2a}[/tex]

The quadratic equation is given below.

x² – 3x – 5 = 0

Then the solution of the equation will be

[tex]\rm x = \dfrac{-(-3)\pm \sqrt{(-3)^2 - 4 *1*(-5)}}{2*1}\\x = \dfrac{3 \pm \sqrt{29}}{2}[/tex]

Then the correct option is B.

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Answer:

∠ABC =   52°

∠BCA =  48°

∠CAB = 80°

Step-by-step explanation:

Angle sum property of triangle: Sum of all the angles of a triangle is 180.

13x² + 12x² + 20x² = 180

Combine like terms,

                 45x²  = 180

                     x² = 180 ÷ 45

                     x² = 4

                     x = √4

                     x = ± 2

x = 2 ; -2

∠ABC = 13x² = 13*4 = 52°

∠BCA = 12x² = 12 *4 = 48°

∠CAB = 20x² = 20 *4 = 80°

The value of x is +2 and -2 and the angles will be given as ∠ABC = 52 ∠ACB = 48 and ∠CAB = 80.

The branch of mathematics sets up a relationship between the sides and the angles of the right-angle triangle is termed trigonometry.

Given that:-

∠ABC = 13x² ∠ACB = 12x² and ∠CAB = 20x²

The angles will be calculated as follows:-

13x² + 12x² + 20x² = 180

45x²  =  180

x²  =  4

x   =  ±2

The angles will be:-

∠ABC = 13x²  = 13 x ( 2² ) = 52

∠ACB = 12x²  = 12 ( 2² )    = 48

∠CAB = 20x²  = 20 ( 2² )  = 80

Therefore value of x is +2 and -2 the angles will be given as ∠ABC = 52 ∠ACB = 48 and ∠CAB = 80.

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Answer: To solve a system of equations by substitution, we can rewrite a two-variable equation as a single variable equation by substituting the value of a variable from one equation into the other. Let’s start by solving the system of equations that we looked at above: x=4 x = 4 y+x=12 y +x = 12

Step-by-step explanation:

Answer:

Isolating a variable in one of the equations.

Step-by-step explanation:

If BCD was reflected over the x axis then the coordinates will be B(-1,0) C(-1,2) and D(-4,1).

Given BCD is a triangle having coordinates (-1,0),(-1,-2),(-4,-1).

When the triangle is reflected over the x axis then all the values of y will change from negative to positive and the values of x remains same because in the second and third quadrant the values of x remains same and in the second quadrant the values of y used to  be positive  and in the third quadrant the values of y used to be negative.

So coordinates of B will change from (-1,0) to (-1,0)

coordinates of C will change from (-1,-2) to (-1,2)

coordinates of D will change from (-4,-1) to (-4,1)

New coordinates will be (-1,0) (-1,2) (-4,1).

Hence if BCD was reflected over x axis the new coordinates will be (-1,0) (-1,2) (-4,1).

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If quadrilateral JKLM were in third quadrant then the coordinates will be (-1,0),(-4,-1),(-4,-2)(-1,-2)

Given Quadrilateral JKLM is in fourth quadrant and coordinates (1,0),(4,-1),(4,-2)(1,-2)

We know that the coordinates of x used to negative in third quadrant and positive in fourth quadrant. Coordinates of y used to negative in third and fourth quadrant.

So in order to change the coordinates from fourth to third we need to change the coordinates of x axis to negative and the coordinates will be :

Coordinates of J will change from (1,0) to (-1,0)

Coordinates of  K will change from(4,-1) to (-4,-1)

Coordinates of L will change from (4,-2) to (-4,-2)

Coordinates of M will change from (1,-2) to (-1,-2).

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