Who can get these correct for me ?

1. has solution x=-2 and goes in the left circle

2. has solutions x=-2 and x=5 and goes in the intersection between the two circles

3. has solution x=-2 and goes in the left circle

4. has solution x=5 and goes in the right circle

Step-by-step explanation:

Solve each equation by factoring

2x^2-12x-32=0

2(x^2-6x-16)=0

2(x+2)(x-8)

x=-2,x=8

Therefore, 1. has solution x=-2 and goes in the left circle

2x^2-6x-20=0

2 (x^2-3x-10) = 0

2(x-5)(x+2)=0

x=-2,x=5

Therefore, 2. has solutions x=-2 and x=5 and goes in the intersection between the two circles

3x^2+2x-8=0

(x+2)(3x-4)=0

x=-2, x=4/3

Therefore, 3. has solution x=-2 and goes in the left circle

3x^2-18x+15=0

3(x^2-6x+5)=0

3(x-1)(x-5)=0

x=1,x=5

Therefore 4. has solution x=5 and goes in the right circle

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step-by-step explanation:

, and

for

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote , , , so that we can call

note the following:

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from , to get to , we add

from , to get to , we add

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

, and for