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SSSSS [86.1K]
3 years ago
9

Why is 1/6 greater than 1/8 but less than 1/3

Mathematics
2 answers:
maksim [4K]3 years ago
7 0
If you change all the fractions so they have a common denominator, you get that 1/6 becomes 4/24, 1/8 becomes 3/24, and 1/3 becomes 8/24. 4/24 is more than 3/24. 4/24 is less than 8/24.
VladimirAG [237]3 years ago
5 0

So first thing we have to realise is that we have to make them have equal denominators. So,

\frac{1}{6} =  \frac{4}{24}  \\ \\   \frac{1}{8} =  \frac{3}{24}  \\  \\  \frac{1}{3} =  \frac{8}{24}.

Now order the fractions....so,

\frac{1}{3} = \frac{8}{24}  \\  \\ \frac{1}{6} = \frac{4}{24} \\ \\ \frac{1}{8} = \frac{3}{24}

Now we can see why \frac{1}{6} is greater than \frac{1}{8} but less than \frac{1}{3} 

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Write the equation of a line that passes through two points (-1, 0) and (4, -5).
sleet_krkn [62]

Answer:

<u>y = -1 (x) -1</u>

Step-by-step explanation:

If you plug in the numbers from the coordinates into the equation, you will find that it's true.

-1 is x and 0 is y. Plug in the #s like so:

y = -1 (x) -1

0 = -1 (-1) - 1

0 = 1 - 1

0 = 0

Therefore, the equation above represents the line that passes through both these points.

8 0
4 years ago
The amount of money in a savings account after t years is represented by the function f(t)=3600(1.035)t .
timama [110]

Answer:  The initial amount in the account = $3600

Step-by-step explanation:

Since we have given that

The amount of money in a savings account after t years is represented by the function

f(t)=3600(1.035)^t

As we know the formula "Compound Interest" :

Amount=P(1+\frac{r}{100})^t\\\\Here,\\\\\text{ P denotes principal}\\\\\text{ R denotes Rate of interest}\\\\\text{ T denotes time}

So, According to our question,

Rate of interest = 0.35 = 35%

So, equate the both the equations , we get that

Hence, The initial amount in the account = $3600

5 0
3 years ago
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Answer:

Therefore, the volume of sphere is 1436.03 cubic units.

Step-by-step explanation:

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Attached the solution and work.

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B) Isolate the variable through inverse operations

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