Answer: 54$
Explanation:
Total cost = 16 + 16 + 22
Total cost = 32 + 22
Total cost = 54
Divide each percentage to find a multiplier:
34/3.2 = 10.625
The Dead Sea is 10.625 times saltier than the Indian Ocean.
So if Indian Ocean bucket = 56 grams of salt
Multiply that by our multiplier.
56 * 10.625 = 595 grams
The answer is 595 grams
Answer:
Yes.. you are correct!
Step-by-step explanation:
Please give me brainliest.
Answer:
<em>LCM</em> = 
Step-by-step explanation:
Making factors of 
Taking
common:

Using <em>factorization</em> method:

Now, Making factors of 
Taking
common:

Using <em>factorization</em> method:

The underlined parts show the Highest Common Factor(HCF).
i.e. <em>HCF</em> is
.
We know the relation between <em>LCM, HCF</em> of the two numbers <em>'p' , 'q'</em> and the <em>numbers</em> themselves as:

Using equations <em>(1)</em> and <em>(2)</em>:

Hence, <em>LCM</em> = 
Answer:
its c
Step-by-step explanation: