Answer:
i would add
Step-by-step explanation:
Alright, so we can start by dividing -2 from both sides, getting |5y-1|=20. Then, since 5y-1 is in an absolute value, 5y-1 is either 20 or -20.
In 5y-1=20, we can add one to both sides, getting 5y=21 and y=21/5. In
5y-1=-20, we can add one again, getting 5y=-19 and y=-19/5.
If you have any more equations, make sure to plug both numbers in to check, but otherwise y has two answers , which are -19/5 and 21/5.
If the 3’s were exponents then... i got 50r3
how i solved:
39r3 + 11r3 + 45s3 - 5s3 - 40s3
50r3 + 40s3 - 40s3
50r3 + 0
= 50r3
Answer: order 2
Step-by-step explanation:
Answer:
1=a4+3
Step 1: Subtract a^4+3 from both sides.
1−(a4+3)=a4+3−(a4+3)
−a4−2=0
Step 2: Add 2 to both sides.
−a4−2+2=0+2
−a4=2
Step 3: Divide both sides by -1.
−a4
−1
=
2
−1
a4=−2
Step 4: Take root.
a=±(−2)(
1
4
)
Answer:
No real solutions.
