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insens350 [35]
3 years ago
9

What makes 2.4=6x true?

Mathematics
1 answer:
Murrr4er [49]3 years ago
4 0
2.4=6x so isolated the variable, 2.4/6=0.4 so x=0.4
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Robert plans to make a box-and-whisker plot of the following set of data. 27, 14, 46, 38, 32, 18, 21 Which of the following shou
Simora [160]
To make a box and whisker plot, you need to draw a number line that would be able to represent all the values in the data (look at your lowest and highest numbers).

You will need to graph the following 5 points on your number line (median, lower quartile, lower extreme, upper quartile, and upper extreme).

Create a box around the quartiles and extend the lines to the extremes.
See the picture I have attached for the box and whisker box for this data set.

7 0
3 years ago
How do I work this out oof
kakasveta [241]

I am so sorry for giving you the wrong answer at first.

Here's the processs

1x=-0.5555

Move the decimal point in the repeating decimal one space forward

10x=-5.55555

Then subtract

10x=5.5555

x=-0.5555= 9x=-5

Divide both sides by 9

9x=-5

9x/9=-5/9

-5/9 is your final answer!

Hope this helps!

7 0
3 years ago
Read 2 more answers
A regular pentagon has a perimeter of 20 cm
Nikolay [14]
Pentagons have five sides, so 5x = 20. Only value for x that makes this true is 4, so 4 cm per side.
4 0
3 years ago
Read 2 more answers
Determine whether each of the binary relation R defined on the given sets A is reflexive, symmetric, antisymmetric, or transitiv
ki77a [65]

Answer:

In explanation

Please let me know if something doesn't make sense.

Step-by-step explanation:

a)

*This relation is not reflexive.

0 is an integer and (0,0) is not in the relation because 0(0)>0 is not true.

*This relation is symmetric because if a(b)>0 then b(a)>0 since multiplication is commutative.

*This relation is transitive.

Assume a(b)>0 and b(c)>0.

Note: This means not a,b, or c can be zero.

Therefore we have abbc>0.

Since b^2 is positive then ac is positive.

Since a(c)>0, then (a,c) is in R provided (a,b) and (b,c) is in R.

*The relation is not antisymmretric.

(3,2) and (2,3) are in R but 3 doesn't equal 2.

b)

*This relation is reflective.

Since a^2=a^2 for any a, then (a,a) is in R.

*The relation is symmetric.

If a^2=b^2, then b^2=a^2.

*The relation is transitive.

If a^2=b^2 and b^2=c^2, then a^2=c^2.

*The relation is not antisymmretric.

(1,-1) and (-1,1) is in the relation but-1 doesn't equal 1.

c)

*The relation is reflexive.

a/a=1 for any a in the naturals.

*The relation is not symmetric.

Wile 4/2 is an integer, 2/4 is not.

*The relation is transitive.

If a/b=z and b/c=y where z and y are integers, then a=bz and b=cy.

This means a=cyz. This implies a/c=yz.

Since the product of integers is an integer, then (a,c) is in the relation provided (a,b) and (b,c) are in the relation.

*The relation is antisymmretric.

Assume (a,b) is an R. (Note: a,b are natural numbers.) This means a/b is an integer. This also means a is either greater than or equal to b. If b is less than a, then (b,a) is not in R. If a=b, then (b,a) is in R. (Note: b/a=1 since b=a)

6 0
3 years ago
Solve:<br> -4 &lt;-2x + 2 &lt; 10
kvv77 [185]

Answer:

3 > x > − 4 and your welcome bud

4 0
3 years ago
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