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Dafna1 [17]
3 years ago
8

What is the equation for the line of reflection A. X=6 B. Y=6 C. Y=x D. Y=2

Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
5 0

Answer:

x = 6

Step-by-step explanation:

The line of reflection is x = 6.

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Since opening night, attendance at play A has increased steadily, while attendance at play B first rose and then fell. Equations
lina2011 [118]
The two lines intersect at the point (4, 214). This means that the attendance was 214 for both plays on the 4th night. 

<span>B. The attendance was the same on day 4. The attendance was 214 at both plays that day.</span>
3 0
2 years ago
Read 2 more answers
Does anyone know the answer to this question.....0.5x + 1.75 = 10
marysya [2.9K]

Answer:

x = 16.5

Step-by-step explanation:

0.5x + 1.75 = 10

Subtract 1.75 from both sides:

0.5x = 8.25

Divide both sides by 0.5

x = 16.5

Check your work!

0.5(16.5) + 1.75 = 10

8.25 + 1.75 = 10

10 = 10

x = 16.5

3 0
3 years ago
Write an equation of the line that passes through (1,-3) and is perpendicular to the line <img src="https://tex.z-dn.net/?f=y%3D
Lubov Fominskaja [6]

Answer:

y = 3x

Step-by-step explanation:

Given line slope = m = -1/3

   perpendicular line slope = - 1/m = 3

Point slope form through point 1, - 3

(y+3) = 3 ( x-1)       expand and rearrange

y -3 = 3x- 3

  y = 3x - 6  

3 0
1 year ago
3 determine the highest real root of f (x) = x3− 6x2 + 11x − 6.1: (a) graphically. (b) using the newton-raphson method (three it
Juliette [100K]

(a) See the first attachment for a graph. This graphing calculator displays roots to 3 decimal places. (The third attachment shows a different graphing calculator and 10 significant digits.)

(b) In the table of the first attachment, the column headed by g(x) gives iterations of Newton's Method. (For Newton's method, it is convenient to let the calculator's derivative function compute the derivative f'(x) of the function f(x). We have defined g(x) = x - f(x)/f'(x).) The result of the 3rd iteration is ...

... x ≈ 3.0473167

(c) The function h(x₁, x₂) computes iterations using the secant method. The results for three iterations of that method are shown below the table in the attachment. The result of the 3rd iteration is ...

... x ≈ 3.2291234

(d) The function h(x, x+0.01) computes the modified secant method as required by the problem statement. The result of the 3rd iteration is ...

... x ≈ 3.0477377

(e) Using <em>Mathematica</em>, the roots are found to be as shown in the second attachment. The highest root is about ...

... x ≈ 3.0466805180

_____

<em>Comment on these methods</em>

Newton's method can have convergence problems if the starting point is not sufficiently close to the root. A graphing calculator that gives a 3-digit approximation (or better) can help avoid this issue. For the calculator used here, the output of "g(x)" is computed even as the input is typed, so one can simply copy the function output to the input to get a 12-significant digit approximation of the root as fast as you can type it.

The "modified" secant method is a variation of the secant method that does not require two values of the function to start with. Instead, it uses a value of x that is "close" to the one given. For our purpose here, we can use the same h(x1, x2) for both methods, with a different x2 for the modified method.

We have defined h(x1, x2) = x1 - f(x1)(f(x1)-f(x2))/(x1 -x2).

6 0
2 years ago
An animal shelter has a bin filled with 456 pounds of dog food there are 4 large dogs at the shelter who each eat 2 pounds of th
Y_Kistochka [10]

If there are 4 dogs and they each eat 2 pounds of food each day. Then, all together there are 8 pounds of food getting eaten each day. You will want to divide 456 by 8.

456/8= 57 days.

456 punds of dog food will last 57 days for 4 dogs.

7 0
3 years ago
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