Question

What is the solution to the following system?
-3x - y = -3
9x + 3y = 9

(3, 3)

(3, -3)

no solution

infinite solutions on the line

Answer 1

infinite

Answer 2

yes the last option

Answer 3

can u guys help me with this one

Answer 4

to clear fractions from the equation 5/6n - 1/3n = -15 you could multiply each term on both sides by the number 6. True False

Answer 5

yes true

Answer 6

If a line has a slope of 2/5 then a perpendicular line would have a slope of-5/2 true false

Answer 7

true

Answer 8

If you were asked to define 3 consecutive odd integers, you would name them x, x + 2, x + 4.

True
False

Answer 9

true

Answer 10

infinite

Answer 11

The polynomial 15xy+3x3y4 is what degree?

6

5

7

9

Answer 12

5

Answer 13

The graph of the solution to the inequality -5x – 12 > 3 has an open circle at -3 and is shaded to the left.

True
False

Answer 14

??

Answer 15

true true true

Answer 16

What is the slope of the line for the equation y = 3?

3

0

undefined

none of the above

Answer 17

0

Answer 18

Solve the absolute value equation |2n + 15| = 1. One solution is n = -7. What is the other solution?

n = -14

n = 8

n = 7

none of the above

Answer 19

??

Answer 20

none

Answer 21

the degree of the polynomial 15xy +3x3y4 is 7, not 5.

Answer 22

thats the power i thought that was coefficient
so its 7 you are right

Answer 23

the graph of the ... is an open circle ... TRUE

Answer 24

What is the slope of the line 2x – 5y = -10

Answer 25

2/5

Answer 26

Identify the x and y intercepts for the equation 4x – 3y = -12.

(4, 0) and (0, -3)

(3, 0) and (0, -4)

(4, 0) and (0, 3)

(-3, 0) and (0, 4)

(3, 0) and (0, 4)

Answer 27

?

Answer 28

(-3, 0) and (0, 4)

Answer 29

the fourth one, (-3,0) and (0,4)

Answer 30

how good are u at this math stuff ?

Answer 31

What type of polynomial is 3x2+6x+1?

monomial

binomial

trinomial

not a polynomial

Answer 32

do we have to answer the other question? LOL

Answer 33

trinomial

Answer 34

lol why

Answer 35

next question, pls. hihi