A times 34 idfbutcgnjytfxvhitd
Answer:
general rule Un=1.8n-21
U100=1.8(100)-21=159
Step-by-step explanation:
a5=a+4d, a5=-12
a10=a+9d, a10=-3
-12=a+4d
-3=a+9d subtract the system of equations
-9=-5d solve for d
d=1.8
substitute d into first equation to find a
-12=a+4(1.8)
a=-19.2
un. = a + (n − 1)d
Un=-19.2+(n-1)(1.8)
Un=1.8n-21
Answer:
<em>Riley has 72 tokens</em>
Step-by-step explanation:
<u>System of Equations
</u>
We have two conditions for the tokens Riley and Erik have earned. Let's call x and y to the number of tokens of Riley and Erik respectively. The first condition states that

Solving for y

The second condition is that the ratio of the number of tokens that Riley had to the number of tokens that Erik has is 8 to 7. It's written as

Or equivalently

Replacing y from the first equation

Operating

Simplifying


Riley has 72 tokens
Solving
we get, 
Step-by-step explanation:
We need to solve: 
Solving:

Adding -ln(2) on both sides:


Using logarithmic rule: if 
So,

Simplifying:

So, Solving
we get, 
Keywords: Logarithms
Learn more about Logarithms at:
#learnwithBrainly
Answer:
for the second questio it is certainly The last choice D
Step-by-step explanation: