Answer:
x<_2
Step-by-step explanation:
first off solve this like a regular equation the >_ dosent matter till later
so basically first multiply - by 13 and -x so you get -13 and x (multiplying two negatives = a positive)
so its 5x-13+x>_9x-7
add 5x and x to get 6x-13 on one side
then you get 6x-13>_9x-7 then subtract -9x from both sides
so its now -3x-13>_-7 add 13 so you get -3x>_6
then divide both sides by -3 (when you divide an equation with a >< or <_ >_ by a negative number that sign switches) so it will now be x<_2
that your final answer hope this helps :)
Answer:
1
Step-by-step explanation:
Numbers to the "right of 0" implies the positive numbers. And an integer has no fractional component. Thus, the first integer to the right of 0 would be 1.
Cheers.
Answer:
50
Step-by-step explanation:
[15 ÷ 5 • 3 + (2³ – 3)] + [4 • (36 – 3³)]
[3 × 3 + (8 - 3)] + [4 × (36 - 27)]
(9 + 5) + (4 × 9)
14 + 36
50
Answer:
61 - 10x
2k
42 / 6 = 7
38 + 2m
3y - 6
Step-by-step explanation:
A. Sixty-one diminished by ten x
= 61 - 10x
B. The product of twice a number k
= 2 * k
= 2k
C. The quotient of forty-two and six is seven
42 / 6 = 7
D. Twice the total of nineteen and a number m.
= 2(19 + m)
= 38 + 2m
E. Six less than 3 times y
3y - 6