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3 years ago

1/8 divided by 11/4 in a fration

Mathematics

1 answer:

lisabon 2012 [21]3 years ago

5 0

Answer:

The answer is 1/22.

Step-by-step explanation:

→ (1/8) ÷ (11/4)

→ (1/8) × (4/11)

→ (1×4)/(8×11)

→ 4/88

→ 1/22 [Simplest form]

Thus, 1/22 is the solution.

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Kayla paints a bookcase she uses 1 5/6 cups of paint on the outside of the bookcase and 3/8 cup of paint on the inside. How many

olchik [2.2K]

Answer: Kayla used 2 5/24 cups of paint all together.

Step-by-step explanation:

Kayla paints a bookcase by using

1 5/6 cups of paint on the outside of the bookcase. Converting 1 5/6 cups to improper fraction, it becomes 11/6 cups of paint.

She also used 3/8 cup of paint on the inside.

Therefore, the total number of cups of paint that Kayla used all together would be

11/6 + 3/8 = (44 + 9)/24 = 53/24 cups of paint. Converting to mixed fraction, it becomes

2 5/24 cups of paint

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3 years ago

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Vladimir79 [104]

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3 years ago

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The function f(x) = (10)–x is reflected across the x-axis to create the function g(x). Which ordered pair is on g(x)? (–2, 75) (

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3 years ago

What is the area of the composite figure shown.

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3 years ago

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

7 0

3 years ago