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dem82 [27]
3 years ago
10

1.) How many one-thirds are in seven halves?

Mathematics
2 answers:
Ilia_Sergeevich [38]3 years ago
4 0

Answer:

1) Twenty-one halves of one-thirds are in seven halves.

2) Fourteen-thirds of three-fourths are in five halves.

Step-by-step explanation:

1) To find : How many one-thirds are in seven halves?

Solution :

Let the required number be 'x',

According to question,

\frac{1}{3}\times x=\frac{7}{2}

Solving the equation,

x=\frac{7\times 3}{2}

x=\frac{21}{2}

Therefore, Twenty-one halves of one-thirds are in seven halves.

2) To find : How many three-fourths are in five halves?

Solution :

Let the required number be 'x',

According to question,

\frac{3}{4}\times x=\frac{5}{2}

Solving the equation,

x=\frac{7\times 4}{2\times 3}

x=\frac{14}{3}

Therefore, Fourteen-thirds of three-fourths are in five halves.

blsea [12.9K]3 years ago
3 0
1/3 in 7/2

2/6 in 21/6

approximately 10 and 1/6 more

3/4 in 5/2

3/4 in 10/4

approximately 3 and 1/4

hope this helps
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Answer:

n/5 = 18

Step-by-step explanation:

Quotient means division.

n/5 = 18

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How many different ways are there to pick a set of 5 crayons from a box of 8 crayons?
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The answer would be 56.
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If x = –3 is the only x-intercept of the graph of a quadratic equation, which statement best describes the discriminant of the e
natita [175]

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If a quadratic equation only has one root, that's a repeated root corresponding to a discriminant of zero.

In this example our equation is something like 2(x+3)^2=0, or expanded

2x^2+12x + 18 = 0

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7 0
3 years ago
What is 04.151515... into a fraction
Sergio039 [100]

Answer:

\displaystyle 4.\overline{15} = \frac{137}{33}.

Step-by-step explanation:

Start by separating this decimal number into its integer part and its fraction part:

4.151515\cdots = 4 + 0.151515\cdots

The most challenging task here is to express 0.151515\cdots as a proper fraction. Once that fraction is found, expressing the original number 4.151515\cdots will be as simple as rewriting a mixed number as an improper fraction.

Let x = 0.151515\cdots. (x + 4) would then represent the original number.

Note that the repeating digits appear in groups of two. Therefore, if the digits in x are shifted to the left by two places, the repeating part will continue to match:

\begin{aligned}x = 0.&151515\cdots && \\ 100\, x = 15.& 151515\cdots \end{aligned}.

Note, that this "shifting" is as simple as multiplying the initial number by 100 (same as 10 raised to the power of the number of digits that needs to be shifted.)

Subtract the original number from the shifted number to eliminate the fraction part completely:

\begin{aligned}&(100\, x) - x \\ &= 15.151515\cdots\\  & \phantom{=}- 0.151515\cdots\\&=15 \end{aligned}.

In other words:

99\, x = 15.

\displaystyle x = \frac{15}{99} = \frac{5}{33}.

Therefore, the original number would be:

\displaystyle x + 4 = \frac{5}{33} = \frac{132 + 5}{33} = \frac{137}{33}.

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