Answer:the answer is 18
Step-by-step explanation:You do 9minus three then get six and then multiply six by three to get 18.
Answer:
D
Step-by-step explanation:
using the rule of logarithms
log x - log y ⇔ log (
)
then
(
) =
a -
d
Answer:
The estimated sum is ≈10
Step-by-step explanation:
This depends on how your professor wants you to estimate sums. Logically, though, a value of nearly 4 and a value of just over 6 add to about 10. The exact answer would be a bit over 10, at ≈10.14117
Answer: 26 pieces of $1 bill and 34 pieces of $5 bill
Total Deposit = $196
Total Bills = 60 pieces
Value of $1 = $1
Value of $5 = $5
Lets assume all the 60 pieces were $1 bill.
The total deposit will be 60 x $1 = $60
But we know that the total deposit is $196. So the extra value must have come from the $5.
$196 - $60 = $136
There is still an outstanding amount of $136 not accounted for.
$5 - $1 = $4
The difference in the two bills is $4
Now we can find out the number of $5 bill by dividing the extra $136 by 4:
136 ÷ 4 = 34 pieces
There are 34 pieces of $5 bill.
Since there are 34 pieces of $5, the rest must be the $1 bill
60 - 34 = 26
There are 26 pieces of $1 bill.
------------------------------------------------------------------------------------
Answer: 26 pieces of $1 bill and 34 pieces of $5 bill.
------------------------------------------------------------------------------------
Answer:
See Explanation
Step-by-step explanation:
The question is incomplete as the number line is not attached.
<em>However, the following will guide you through</em>
Given

Required
Determine BC
First, we need to get the fraction of BC using the following

Substitute
for AB

Make BC the subject of formula

Take LCM



This implies that you multiply the length of AC by 2/3 to get BC
<em>Take for instance, AC = 24</em>
<em>BC will be:</em>
<em />
<em />
<em />
<em />
<em />
<em />
<em />
<em>Or if AC = 12</em>
<em>BC will be:</em>
<em />
<em />
<em />
<em />
<em />
<em />
<em />
<em>Or if AC = 36</em>
<em>BC will be:</em>
<em />
<em />
<em />
<em />
<em />
<em />
<em />