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Mashcka [7]
3 years ago
13

How to solve this problem?

Mathematics
1 answer:
SOVA2 [1]3 years ago
6 0
Covert to a common denominator to help:
A LCD is 24.
12/24, 20/24, and 15/24

So 12/24, 15/24, 20/24
Convert back into their original denominators:
1/2, 5/8, 5/6.
F. 
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Decide whether the change is an increase or a decrease, and find the percent change. Original number = 35; New number = 16.... P
Elodia [21]

Answer: 60 decrease or 150 decrease

Step-by-step explanation:

3 0
3 years ago
Show me subtraction fractions with unlike denominator 5 grad super teacher work sheets
lawyer [7]
1/3-1/5     

5/3-3/2    

I hoped this helped
5 0
3 years ago
Three times the sum of a number and 15 is 24
allsm [11]

Answer:

\huge\boxed{\sf x = -7}

Step-by-step explanation:

<em>Let the number be x </em>

So,

=> The sum of number and 15 is x + 15

=> Three times will be 3(x+15)

<u>Given that</u> it equals 24

<u>So, the given condition is:</u>

3(x+15) = 24

<em>Resolving Parenthesis</em>

3x + 45 = 24

<em>Subtracting 45 to both sides</em>

3x = 24 - 45

3x = -21

<em>Dividing both sides by 3</em>

x = -7

7 0
3 years ago
Read 2 more answers
Given that 6.00 Ghanaceasa dolar
mash [69]

Answer:

do you have options to this

4 0
3 years ago
If 150 grams of a radioactive isotope are present at 2:00 PM and 10 grams remain at 6:00 PM (the same day), what is the half-lif
pogonyaev

Initial amount, A_o=150\ g .

Final amount, A =10\ g .

Time taken, t = 6:00 - 2:00 = 4 hour.

We know,

A=A_o(\dfrac{1}{2})^{\dfrac{t}{h}}\\\\10 = 150 \times \dfrac{1}{2}^{\dfrac{4}{h}}\\\\2^{\dfrac{4}{h}}=15\\\\\dfrac{4}{h}= log_215\\\\h = \dfrac{4}{log_215}\\\\h = 1.024 \ hours

Therefore, the half-life of the isotope is 1.024 hours.

Hence, this is the required solution.

8 0
3 years ago
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