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Damm [24]
3 years ago
15

5x + y = 11 3x − y = 9

Mathematics
1 answer:
siniylev [52]3 years ago
6 0

Answer:

  • x = 5/2
  • y = -1.5

Step-by-step explanation:

  • 5x + y = 11

        3x − y = 9

  • => 5x + y = 11

           +3x − y = +9

  • => 8x = 20

  • => x = 20/8

  • => x = 5/2

  • => 5(5/2) + y = 11

  • => 25/2 + y = 11

  • => 12.5 + y = 11

  • => y = 11 - 12.5

  • => y = -1.5

Therefore, the answers are:

  • x = 5/2
  • y = -1.5

Hoped this helped.

BrainaicUser1357

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Without calculating, which has a bigger volume. A cube that has a length, width, and height of 18 m. Or a sphere with a radius o
Evgesh-ka [11]

Weird. A period appears above this... huh.

Answer:

[Th]e cube has a greater value.

Step-by-step explanation:

What the word problem really wants us to get [is ]the question of 'Which is greater, A=6a^2 when 'a' [is] 18 or A=4\pir^2 when r = 9? And here's how to solve that.

Starting with the[ c]ube we have A=6a^2. A bit t[o]o simple, right?

A=6(18)^2 Substitute numbers.

A=6(324) Solve ex[p]onents.

A=1944 Mult[i]ply.

So w[e] know that the cube is 1944 meters cube[d ] in area. But what about the more [f]ormidable sphere? Fo[r] it we need a slightly m[o]re co[m]plicated formula, A=4\pir^2. However, instead of using the real pi I will be rounding to 3.14, since we have no calculator so anything more would take way too long and fry your[ bra]in.

A=4(3.14)(9)^2 Subst[i]tute numbers.

A=4(3.14)(81) Solve expone[n]ts.

A=12.56(81) Multip[ly].

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Now, since I'm sure all of us can count, we know that 1944 is greater than 1017.36. Or in other words, the cube is bigger than the sphere.

And PLEASE don't copy this guys. Make your own iteration. Change it up!

3 0
3 years ago
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Volgvan

Answer:

The numerator must be greater than 16

Step-by-step explanation:

We are told that a certain fraction is greater than 2. Thus, if the numerator and denominator are x and y respectively, then we have;

x/y > 2

We are further told that the denominator is 8.

Thus;

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Multiply both sides by 8 to give;

x > 16

Thus,the numerator must be greater than 16.

5 0
3 years ago
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