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Vinil7 [7]
3 years ago
6

PLEASE ANSWER QUICK ITS A TIMED QUIZ Part A: Explain why the x-coordinates of the points where the graphs of the equations y = 4

^x and y = 2^x−2 intersect are the solutions of the equation 4^x = 2^x−2. (4 points) Part B: Make tables to find the solution to 4^x = 2^x−2. Take the integer values of x between −3 and 3. (4 points) Part C: How can you solve the equation 4^x = 2^x−2 graphically? (2 poin
Mathematics
1 answer:
kramer3 years ago
6 0

A.  We have two lines:  y = 4-x   and   y = 8-x^-1

Given two simultaneous equations that are both to be true, then the solution is the points where the lines cross. The intersection is where the two equations are equal. Therefore the solution that works for both equations is when

4-x = 8-x^-1

This is where the two lines will cross and that is the common point that satisfies both equations.

 

B.  4-x = 8-x^-1

 

 x         4-x      8-x^-1

______________

 

-3          7        8.33

-2          6        8.5

-1          5        9

 0          4        -

 1          3        7

 2          2        7.5

 3          1        7.67

 

The table shows that none of the x values from -3 to 3 is the solution because in no case does

4-x = 8-x^-1

 

To find the solution we need to rearrange the equation to find for x:

4-x = 8-x^-1

Multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

 

Therefore there are two points that satisfies the equation.

Find y:  

x=-4.236

y = 4-x  = 4 – (-4.236) = 8.236

y = 8-x^-1 =  8-(-4.236)^-1 = 8.236

 

x=0.236

 y = 4-x  = 4 – (0.236) = 3.764

y = 8-x^-1 =  8-(0.236)^-1 = 3.764

 

Thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

 

C.  To solve graphically the equation 4-x = 8-x^-1

We would graph both lines: y = 4-x  and   y = 8-x^-1

The point on the graph where the lines cross is the solution to the system of equations.

Just graph the points on part B on a Cartesian coordinate system and extend the two lines.  The solution is, as stated, the point where the two lines cross on the graph.

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Quadrilateral PEST has vertices (-1, -5), (8, 2), (11, 13), and (2, 6), respectively. Classify the quadrilateral as a square, rh
Alchen [17]

Answer:

The figure PEST is a rhombus

Step-by-step explanation:

* Lets talk about the difference between all these shapes

- At first to prove the shape is a parallelogram you must have one

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# Each two opposite sides are parallel OR

# Each two opposite sides are equal in length OR

# Its two diagonals bisect each other

- After that to prove the parallelogram is:

* A rectangle you must have one of these conditions

# Two adjacent sides are perpendicular to each other OR

# Its two diagonals are equal in length

* A rhombus you must have one of these conditions

# Two adjacent sides are equal in length OR

# Its two diagonals perpendicular to each other OR

# Its diagonals bisect its vertices angles

* A square you must have two of these conditions

# Its diagonals are equal and perpendicular OR

# Two adjacent sides are equal and perpendicular

* Now lets solve the problem

∵ The vertices of the quadrilateral PEST are

   P (-1 , -5) , E (8 , 2) , S (11 , 13) , T (2 , 6)

- Lets find the slope from each two points using this rule :

 m = (y2 - y1)/(x2 - x1), where m is the slope and (x1 , y1) , (x2 , y2)

 are two points on the line

- Let (x1 , y1) is (-1 , -5) and (x2 , y2) is (8 , 2)

∴ m of PE = (2 - -5)/(8 - -1) = 7/9

- Let (x1 , y1) is (8 , 2) and (x2 , y2) is (11 , 13)

∴ m of ES = (13 - 2)/(11 - 8) = 11/3  

- Let (x1 , y1) is (11 , 13) and (x2 , y2) is (2 , 6)

∴ m of ST = (6 - 13)/(2 - 11) = -7/-9 = 7/9

- Let (x1 , y1) is (2 , 6) and (x2 , y2) is (-1 , -5)

∴ m of TP = (-5 - 6)/(-1 - 2) = -11/-3 = 11/3

∵ m PE = m ST = 7/9

∴ PE // ST ⇒ opposite sides

∵ m ES = m TP = 11/3

∴ ES // TP ⇒ opposite sides

- Each two opposite sides are parallel

∴ PEST is a parallelogram

- Lets check if the parallelogram can be rectangle or rhombus or

 square by one of the condition above

∵ If two line perpendicular , then the product of their slops = -1

- Lets check the slopes of two adjacent sides (PE an ES)

∵ m PE = 7/9

∵ m ES = 11/3

∵ m PE × m ES = 7/9 × 11/3 = 77/27 ≠ -1

∴ PE and ES are not perpendicular

∴ PEST not a rectangle or a square (the sides of the rectangle and

  the square are perpendicular to each other)

- Now lets check the length of two adjacent side by using the rule

 of distance between two points (x1 , y1) and (x2 , y2)

 d = √[(x2 - x1)² + (y2 - y1)²]

- Let (x1 , y1) is (-1 , -5) and (x2 , y2) is (8 , 2)

∴ PE = √[(8 - -1)² + (2 - -5)²] = √[9² + 7²] = √[81 + 49] = √130 units

- Let (x1 , y1) is (8 , 2) and (x2 , y2) is (11 , 13)

∴ ES = √[(11 - 8)² + (13 - 2)²] = √[3² + 11²] = √[9 + 121] = √130 units

∴ PE = ES ⇒ two adjacent sides in parallelogram

∴ The four sides are equal

* The figure PEST is a rhombus

4 0
2 years ago
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