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

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.

step-by-step explanation:

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!