What would the graph of 10x + 7y < 49 look ?

The graph is attached.

Step-by-step explanation:

1. Solve for y, as following:

2. The equation of the line in slope intercept form is:

Where m is the slope and b is the y-intercept.

3. In this case the equation of the line is:

then:

4. Find the x-intercept. Make y=0. Then:

5. Then, plot the line that passes through the points (0,7) and (4.9, 0).

6. The symbol of the inequality is < therefore, the line must be dashed and indicates that the region under the line must be shaded.

Then you obtain the graph attached.

We need to graph the given inequality 10x + 7y < 49.

We can do that into 2 parts:

Part 1:

Graph the line 10x+7y=49

plug any number for x say x=0

10x+7y=49

10(0)+7y=49

0+7y=49

7y=49

y=49/7

y=7

hence it passes through point (0,7)

similarly plug y=0

10x+7(0)=49

10x=49

x=49/10

x=4.9

hence it passes through point (4.9,0)

Graph both points then join them by a straight dotted line because of < sign.

PART 2:

Shade the graph for inequality sign <

use any test point which is not on the given line say (0,0)

plug into original problem

10x + 7y < 49

10(0) + 7(0) < 49

0<49 is true

so shade in direction of test point

Hence final graph looks like:

step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}