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Liula [17]
3 years ago
8

Name five fractions whose values are between 1/8 and 1/5

Mathematics
2 answers:
TiliK225 [7]3 years ago
8 0
The answer is 1/8 bc thats the anwser i got
Volgvan3 years ago
6 0

Answer:

1/5

1. 1/56

2. 1/6

3. 1/65

4. 1/7

5. 1/75

1/8

Step-by-step explanation:

The fractions I gave (1/56, 1/6, 1/65,  1/7, 1/75) are all in between 1/5 and 1/8.

A way to think about this is to add a zero to a denominator of 1/5 and 1/8 and turn them into whole numbers:

1/5 --> 1/50 --> 50

1/8 --> 1/80 --> 80

Then find number between them:

55, 60, 65, 70, 75

Then, turn them into a fraction with a 1 as the numerator and the number as denominators:

1/55, 1/60, 1/65, 1/70, 1/75

This is the process.  These numbers are all in between 1/5 and 1/8.

I hope this helps you!

(P.S., I don't think this works with negative numbers, though)

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The perimeter of this rectangle is 156 cm.
erastova [34]

Answer:

x = 15

Therefore, the length and width are 31 and 47.

Step-by-step explanation:

Perimeter = (2x + 1) + (2x + 1) + (3x + 2) + (3x + 2) = 156

Simplified,

10x + 6 = 156

Subtract 6 from both sides, then divide by 10.

10x = 150

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To find the length and width, substitute 15 (the value of x) into the individual equations.

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3x^2+kx=-3 What is the value of K will result in exactly one solution to the equation?
Scorpion4ik [409]

Answer:

For k = 6 or k = -6, the equation will have exactly one solution.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = (x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

If \bigtriangleup = 0, the equation has only one solution.

In this problem, we have that:

3x^{2} + kx + 3 = 0

So

a = 3, b = k, c = 3

\bigtriangleup = b^{2} - 4ac

\bigtriangleup = k^{2} - 4*3*3

\bigtriangleup = k^{2} - 36

We will only have one solution if \bigtriangleup = 0. So

\bigtriangleup = 0

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k^{2} = 36

k = \pm \sqrt{36}

k = \pm 6

For k = 6 or k = -6, the equation will have exactly one solution.

3 0
2 years ago
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