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3 years ago

Find the Least Common Denominator of the fractions: 3/5 and 27

Mathematics

1 answer:

Flauer [41]3 years ago

3 0

Answer: 5

Step-by-step explanation:

The LCM for the denominators 5 and 1 is 5. So that means the LCD is 5.

3/5 = 3/5 x 1/1 = 3/5

27 = 27/1 x 5/5 = 135/5

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alexgriva [62]

Answer:

x = 20

Step-by-step explanation:

This isn't too difficult once you understand it.

The bottom left angle of the left triangle (on the other side of 110), that angle is congruent to the 70 degrees angle. that angle is 70 degrees. y = 90 degrees, 90 + 70 = 160

180 - 160 = 20

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2 years ago

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igor_vitrenko [27]

Answer

y = 0.2x − 7

Step-by-step explanation:

Find the slope first by subtracting (10, -5) and (5, -6) in slope formula, this will equal 1/5 or 0.2

Then find the y-intercept which is the point that crosses the y-axis, in this case -7 is the y-intercept

Finally format in y=mx + b for slope-intercept form

y=0.2x-7

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2 years ago

The graph below is a portion of a complete graph. Which graph below is the complete graph assuming it is an even function?

Ierofanga [76]

Answer:

Option b

Step-by-step explanation:

A function is even if and only if f(-x) = f(x).

This means that there is the same value of y for $x_0$ and for $-x_0$

Where $x_0$ is any point belonging to the function .

Therefore you must select a graph whose ordered pairs are of the form $(x_0, y_0)$ and $(-x_0, y_0)$

Therefore the correct answer is option b. Assume that each box in the graph equals 1 unit. Then, for x = 2 y = -3 and for x = -2 y = -3. Then it is shown that the function is even.

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3 years ago

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kumpel [21]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

In a group of a hundred and fifty students attending a youth workshop in mombasa, 125 of them are fluent in kiswahili, 135 in en

jek_recluse [69]

Answer:

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

Step-by-step explanation:

Step(i):-

Given total number of students n(T) = 150

Given 125 of them are fluent in Swahili

Let 'S' be the event of fluent in Swahili language

n(S) = 125

The probability that the fluent in Swahili language

$P(S) = \frac{n(S)}{n(T)} = \frac{125}{150} = 0.8333$

Let 'E' be the event of fluent in English language

n(E) = 135

The probability that the fluent in English language

$P(E) = \frac{n(E)}{n(T)} = \frac{135}{150} = 0.9$

n(E∩S) = 95

The probability that the fluent in English and Swahili

$P(SnE) = \frac{n(SnE)}{n(T)} = \frac{95}{150} = 0.633$

Step(ii):-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = P(S) + P(E) - P(S∩E)

= 0.833+0.9-0.633

= 1.1

Final answer:-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

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2 years ago