Answer:
-3r + 15 ---> answer
Step-by-step explanation:
r < 5
You are going to multiply both sides with 3. The reason being is that 3 is a positive number and the equality sign will not change if you use +3.
3r < 15
Now, subtract 15 from both sides, you will get this:
3r < 15
-15 -15
-------------
3r — 15 < 0
Lastly, using the Modulus function, we are going to add a negative sign to the content of our previous step because it's already negative.
So, -3r + 15 is the final solution if r < 5 in the given equation of l3r-15l
C+s=48
2c+4s=134 ⇒ c+2s=67
subtracting the equations, we get, s=67-48=19,
so, c=48-19=29
Answer:
if it's 2/3r^4 heres the answer
Step-by-step explanation:
1.) 2/3 times r raised to the 4th power
2.) the product of 2/3 and r raised to the 4th power
3.) 2/3 multiplied by r raised to the 4th power
4.) 2/3 of r raised to the 4th power
you can pick one of the many choices I've listed!! youre welcome :)
