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docker41 [41]
2 years ago
10

What are three equivalent ratio for 12/14

Mathematics
2 answers:
defon2 years ago
8 0
12/14  can be simplified, and that would be an equivalent ratio.  Divide 12 and 14 both by 2, and then you get 6/7.

Another equivalent ratio is 120/140.  To get this ratio, I merely multiplied both the numerator and denominator of the ratio by 10.

You can get an infinite amount of equivalent ratios for 12/14.  All you have to do is make sure that whatever you multiply one part of the ratio by, you have to multiply the other part of the ratio by the same thing.

A final example is 60/70.  I just multiplied both 12 and 14 by 5.



gulaghasi [49]2 years ago
6 0

For this case we have the following expression:

\frac{12}{14}

What we must do is rewrite this expression in an equivalent way.

To do this, we multiply the numerator and the denominator of the fraction by the same factor.

We have then:

For k = 1/2:

\frac{12}{14}*\frac{\frac{1}{2}}{\frac{1}{2}} = \frac{6}{7}

For k = 2:

\frac{12}{14} *\frac{2}{2} = \frac{24}{28}

For k = 3:

\frac{12}{14} *\frac{3}{3} = \frac{36}{42}

Answer:

Three equivalent ratio for 12/14 are:

\frac{6}{7}

\frac{24}{28}

\frac{36}{42}

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Please Help!!
algol13
When you see questions of this nature, test the individual inequalities and look out for their intersection.

For
y < \frac{2}{3} x
Choose a point in the lower or upper half plane created by the line
y = \frac{2}{3} x
The above line is the one which goes through the origin.

Now testing (1,0) yields,

0 < \frac{2}{3} (1)
That is,

0 < \frac{2}{3}
This statement is true. So we shade the lower half of
y = \frac{2}{3} x

For
y \geqslant - x + 2
We test for the origin because, it is not passing through the origin.

0 \geqslant - (0) + 2
This yields
0 \geqslant 2
This statement is false so we shade the upper half.

The intersection is the region shaded in B. The top right graph
8 0
3 years ago
Which of the following is the number if sides a polygon can have to form a regular tessallation?
Arisa [49]

None of the options is true.

Three Different, Regular tessellation is formed 1. if we join regular polygons of side 3, called equilateral triangle,2. Regular polygon of side 4, called Square, and 3. Regular polygon of side,6 called hexagon,

to form a pattern.

<h3>What is the polygon?</h3>

A polygon is said to be regular if all interior angles and all sides are equal.

The regular polygon of side 3, when joined together can be named: 3.3.3.

The regular polygon of side 4, when joined together can be named 4.4.4.

The regular polygon of side 6, when joined together can be named 6.6.6.

So, a regular tessellation is formed, from a regular polygon having,3,4, and 6 Sides.

None of the sides of the polygon, which is given in the option, can be used to form a regular tessellation.

None of the options is true.

To learn more about the polygon visit:

brainly.com/question/1592456

#SPJ1

6 0
2 years ago
If you flip a coin and roll a
worty [1.4K]

Answer: 1/12

Step-by-step explanation:

Your outcome, 2H, divided by the total size of the output set. Thus gives you probability 1/12

Multiply the probability of Heads P(x=H | coin) = 1/2 . The probability of 2 P(x=2 | Die) = 1/6. Multiply 1/2(1/6) = 1/12.

7 0
2 years ago
Read 2 more answers
The sum of the squares of two consecutive positive integers is 41. Find the two
Komok [63]

We have to present the number 41 as the sum of two squares of consecutive positive integers.

1² = 1

2² = 4

3² = 9

4² = 16

5² = 25

16 + 25 = 41

<h3>Answer: 4 and 5</h3>

Other method:

n, n + 1 - two consecutive positive integers

The equation:

n² + (n + 1)² = 41     <em>use (a + b)² = a² + 2ab + b²</em>

n² + n² + 2(n)(1) + 1² = 41

2n² + 2n + 1 = 41     <em>subtract 41 from both sides</em>

2n² + 2n - 40 = 0     <em>divide both sides by 2</em>

n² + n - 20 = 0

n² + 5n - 4n - 20= 0

n(n + 5) - 4(n + 5) = 0

(n + 5)(n - 4) = 0 ↔ n + 5 = 0 ∨ n - 4 =0

n = -5 < 0 ∨ n = 4 >0

n = 4

n + 1 = 4 + 1 = 5

<h3>Answer: 4 and 5.</h3>
7 0
2 years ago
What is the<br> the distance between a point at (-3,4) and a point at (2,-5)
zaharov [31]
<h2><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>:-</h2>

What is the distance between a point at (-3,4) and a point at (2,-5) ?

<h2><u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u>:-</h2>

Let the two points be A and B .

And, let the distance be P .

Now, the distance P between the point A (-3,4) and the point B (2,-5) is

\bf P \: = \sqrt{(-3 \: - \: 2)^2 \: + \: (4 \: + \: 5)^2}

\implies \bf P \: = \sqrt{(-5)^2 \: + \: (9)^2}

\implies \bf P \: = \sqrt{25 \: + \: 81}

\implies \bf P \: = \sqrt{106}

\implies \bf P \: = \: 10.3

<h3>The distance between a point at (-3,4) and a point at (2,-5) is <u>1</u><u>0</u><u>.</u><u>3</u> . [Answer]</h3>
4 0
2 years ago
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