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2 years ago

Is -5/12 and irrational or rational

Mathematics

2 answers:

oksian1 [2.3K]2 years ago

5 0

Answer:

Rational.

Step-by-step explanation:

Since it can be put into a fraction, negative or positive it is clearly rational. Irrational numbers are numbers we can never know the true value because they repeat forever (pi for example)

just olya [345]2 years ago

5 0

Answer:

rational number

Step-by-step explanation:

hope this helps...yo

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In the following diagram \overline{DE} \parallel \overline{FG}

zloy xaker [14]

Answer:

<x= 31°

Step-by-step explanation:

m<BCA= m<GCJ (vertical angles)

m<BCA= 59° (substitution)

Since line KL is perpendicular to line FG, the angle formed at point B is 90°.

Therefore, m<ABC=90°

m<BAC+m<ABC+m<BCA=180° (sum of triangle)

m<BAC+90°+59°=180° (substitution)

m<BAC+149°=180°

m<BAC=180°-149°

m<BAC=31°

<x=<BAC (vertical angles)

m<x=31° (substitution)

5 0

3 years ago

The hypotenuse of a right triangle measures 19 meters. One leg measures 15 meters. To the nearest tenth of a meter, what is the

yawa3891 [41]

Answer:

11.7

Step-by-step explanation:

A^2+b^2=c^2

x^2+15^2=19^2

6 0

2 years ago

Urgent. There are 12 coloured counters in a bag. The counters are black, white or grey A counter is chosen at random. The probab

Nastasia [14]

Answer:

1/12

Step-by-step explanation:

Needed information

$\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}$

The sum of the probabilities of all outcomes must equal 1

Solution

We are told that the probability that the counter is not black is 3/4.

As the sum of the probabilities of all outcomes must equal 1, we can work out the probability that the counter is black by subtracting 3/4 from 1:

$\sf P(counter\:not\:black)=\dfrac34$

$\implies \sf P(counter\:black)=1-\dfrac34=\dfrac14$

We are told that the probability that the counter is not white is 2/3.

As the sum of the probabilities of all outcomes must equal 1, we can work out the probability that the counter is white by subtracting 2/3 from 1:

$\sf P(counter\:not\:white)=\dfrac23$

$\implies \sf P(counter\:white)=1-\dfrac23=\dfrac13$

We are told that there are black, white and grey counters in the bag. We also know that the sum of the probabilities of all outcomes must equal 1. Therefore, we can work out the probability the counter is grey by subtracting the probability the counter is black and the probability the counter is white from 1:

$\begin{aligned}\sf P(counter\:grey) & = \sf1-P(counter\:black)-P(counter\:white)\\\\ & =\sf 1-\dfrac14-\dfrac13\\\\ & = \sf \dfrac{12}{12}-\dfrac{3}{12}-\dfrac{4}{12}\\\\ & = \sf \dfrac{12-3-4}{12}\\\\ & = \sf \dfrac{5}{12}\end{aligned}$