Write 0.16666 as a fraction

Mathematics

2 answers:

AVprozaik [17]3 years ago

5 0

The answer to this specific question which is “Write 0.16666 as a fraction.” Is 1/6

olchik [2.2K]3 years ago

4 0

This one you can do by multiplying the number by ten, getting 1.666, or 1+0.66. 0.66 is easy to put into a fraction form as 2/3. add the other 3/3 from the whole 1 to get 5/3. From here you either divide by 10 or multiply by 1/10, as KCF will do that anyways if you divide to get 5/30, or 1/6.

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What is the value of 4 2/3 divided by 2/5

Nitella [24]

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6 0

3 years ago

Which of the following expressions are equivalent to (6.2+6.2)+9.4(6.2+6.2)+9.4left parenthesis, 6, point, 2, plus, 6, point, 2,

ELEN [110]

Answer:

6.2 + (6.2 + 9.4) ⇒ A

6.2 - (-6.2 - 9.4) ⇒ E

Step-by-step explanation:

* Lets revise some definitions to solve the problem

- Associative property states that you can re-group numbers and you

will get the same answer [(a + b) + c = a + (b + c)]

- Additive inverse of a number means reverse the sign of the number

the additive inverse of x is -x

- The expression is (6.2 + 6.2) + 9.4

- We want to find which of the following expressions are equivalent

to this expression

- By using associative property we can write the expression as

6.2 + (6.2 + 9.4)

∴ (6.2 + 6.2) + 9.4 = 6.2 + (6.2 + 9.4)

- Multiply (-) by the additive inverse of number is the number itself

If a number is x and its additive inverse is -x, then (-) × (-x) = x

∵ -6.2 is additive inverse of 6.2

∵ (-)(-6.2) = 6.2

∵ -9.4 is additive inverse of 9.4

∵ (-)(-9.4) = 9.4

∴ 6.2 - (-6.2 - 9.4) = 6.2 + 6.2 + 9.4

∴ (6.2 + 6.2) + 9.4 = 6.2 - (-6.2 - 9.4)

* The answers are A and E

8 0

4 years ago

Find the values of c such that the area of the region bounded by the parabola

KATRIN_1 [288]

Solution-

The two parabolas are,

$y=16x^2-c^2 \ and \ y=c^2-16x^2$

By solving the above two equations we calculate where the two parabolas meet,

$So \ 16x^2-c^2 = c^2-16x^2 \Rightarrow 32x^2=2c^2 \Rightarrow x=\frac{1}{4}c$

Given the symmetry, the area bounded by the two parabolas is twice the area bounded by either parabola with the x-axis.

$\therefore Area=2\int_{-c}^{c}y.dx= 2\int_{-c}^{c}(16x^2-c^2).dx\\=2[\frac{16}{3}x^3-c^2x]_{-c}^{ \ c}=2[(\frac{16}{3}c^3-c^3)-(-\frac{16}{3}c^3+c^3)]=2[\frac{32}{3}c^3-2c^3]=2(\frac{26c^3}{3})\\=\frac{52c^3}{3}$