What is the solution to this system of linear equations?
3x – 2y = 14
5x + y = 32
(3, 5)
(6, 2)
(8, –1)
(14, –18)

Answer : (6, 2)

3x – 2y = 14 > Equation 1

5x + y = 32 > Equation 2

We solve the second equation for y

5x + y = 32

y = 32 - 5x ( Subtract 5x on both sides)

Now substitute 32 - 5x for y in the first equation

3x – 2y = 14

3x - 2(32 - 5x) = 14

3x - 64 + 10x = 14 ( combine like terms)

13x - 64 = 14 ( add 64 on both sides)

13x = 78 ( divide by 13 on both sides)

x = 6

Now we plug in x=6 in y= 32 - 5x

y = 32 - 5(6) = 32 - 30 = 2

So solution is (6,2)

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}

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length