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QveST [7]
3 years ago
15

you bought a mobile kit. you read the package that the weight of the entire mobile is 40 oz, each inch of crossbar weights 1 oz.

what is the weight of each shape? justify your reasoning

Mathematics
1 answer:
Nimfa-mama [501]3 years ago
7 0

Answer:

See below

Step-by-step explanation:

<em>Refer to attache</em>d

Since 40 oz is total

<u>The weight of each shape indicated in the picture and calculated as:</u>

  • 40/2 = 20 oz
  • 20/2 = 10 oz
  • 10/2 = 5 oz

As each inch weights 1 oz, there must be 5 of each shapes included in the kit

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3 years ago
X+6y=27<br> 7x-3y=9 por metodo de igualacion
GalinKa [24]

Answer:

(3,4)

Step-by-step explanation:

The system of equations is:

x+6y=27

7x-3y=9.

I looked up "metodo de igualacion". It is basically American for doing substitution.

However, the only difference is you are asked to solve both equations for a variable.

The first equation looks easy to solve for x. So I'm going to solve both equations for x.

x+6y=27

Subtract 6y on both sides:

x     =-6y+27

7x-3y=9

Add 3y on both sides:

7x    =3y+9

Divide both sides by 7:

x     =3/7 y +9/7

So both equations are solved for x.  You want to find when the x's are the same because you are looking for a common amongst the lines given.

So we have

-6y+27=3/7 y  +9/7

I hate the fractions honestly so I'm going to multiply both sides by 7 so they will no longer be for now:

-42y+189=3y + 9

Now add 42y on both sides:

         189=45y+9

Subtract 9 on both sides:

        180=45y

Divide both sides by 45:

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If 4=y, then y=4.

So now once we have obtain 4 for y, we will use one of the equations given along with it to find x. Just choose one. Choose the easier looking one to you.

I like the x=-6y+27 with y=4.

So replace y with giving you:

x=-6(4)+27

x=-24+27

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So the solution is (x,y)=(3,4).

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8 0
3 years ago
Pls answer this is due by tomorrow so I need help ASAP
leonid [27]

5/6 = 0.83333... (Repeating)

2/11 = 0.181818... (Repeating)

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For the number system diagram, from smallest box to larges, it would be natural numbers, whole numbers, and integers, rational numbers, then real numbers. And irrational numbers off to the side, but considering you only have 3 boxes then put the first 3 I mentioned.

It is definetly possible for a number to be both whole and an integer. As long as it's not a fraction, decimal, or negative value.

5 0
3 years ago
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