I need help with this! Factor Quadratic equation. 10g^2+9g+2

(2g + 1)(5g + 2)

Step-by-step explanation:

Given

10g² + 9g + 2

Consider the factors of the product of the g² term and the constant term which sum to give the coefficient of the g- term.

product = 10 × 2 = 20 and sum = 9

The factors are + 5 and + 4

Use these factors to split the g- term

10g² + 5g + 4g + 2 ( factor the first/second and third/fourth terms )

= 5g(2g + 1) + 2(2g + 1) ← factor out (2g + 1) from each term

= (2g + 1)(5g + 2) ← in factored form

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}

(d)

step-by-step explanation:

given: the line in the graph passes through points (5,30) and (6,26).

then, the equation of line is:

now, put y=a(n) and x=n,

hence, option d is correct.