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3 years ago

Is 11/12 a rational or irrational number

Mathematics

1 answer:

Reika [66]3 years ago

6 0

Answer:

Rational

Step-by-step explanation:

A rational number is any number that can be written as a ratio of integers. 11 and 12 are both integers, so 11/12 is rational.

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What is the standard deviation of the data set given below? 4,7, 8, 9, 12

larisa [96]

Answer:

I think 7

Step-by-step explanation:

4 , 8 , 9 and 12 has factors.

7 has no factor.

4= 2×2

8= 2×4

9= 3×3

12= 3×4

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3 years ago

Can you show that this quadrilateral is a parallelogram?

larisa [96]

The diagonals intersect perpendicularly, two consecutive angled are supplementary, and that opposite angles are equal

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3 years ago

The school band collected $2,892 for 480 tickets sold to their spring concert. Adult tickets were $8 each and student tickets we

kakasveta [241]

A + s = 480......s = 480 - a
8a + 5s = 2892

8a + 5(480 - a) = 2892
8a + 2400 - 5a = 2892
8a - 5a = 2892 - 2400
3a = 492
a = 492/3
a = 164 <=== there were 164 adult tickets sold

5 0

3 years ago

Read 2 more answers

Proportional relationships! Need help on!

Jlenok [28]

Direct variation has this equation:

y = kx

where k is the constant of variation

y = -5 ; x = -15

y = kx
-5 = k(-15)
-5/-15 = k
1/3 = k

Choice D. y = 1/3 x

7 0

3 years ago

I need help w/ this!! thank you

Fittoniya [83]

<h3>Answer:</h3>

$\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}$

<h3>Step-by-step explanation:</h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

$\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2 \\\\\bf\implies (2y+3)^2 \leq 1 \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1 \\\\\bf\implies 4y^2+9+12y - 1 \leq 0 \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0 \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup \\\\\bf\implies y^2y+2y+y+2 \leq 0 \\\\\bf\implies y(y+2)+1(y+2)\leq 0 \\\\\bf\implies ( y+2)(y+1)\leq 0 \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}$

Let's plot a graph to see its interval . Graph attached in attachment .

Now we can see that the Interval notation of would be ,

$\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}$

<h3>Hence the standard form of inequality is y²+3y +2 ≤ 0 and the Solution set of the inequality is [ -2 , -1 ] .</h3>

4 0

2 years ago