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Dmitrij [34]
2 years ago
12

12gfdyhgvkjhfvkuhj ,jvhgctkgdcfmtghjm

Mathematics
1 answer:
lesantik [10]2 years ago
3 0

Answer:

1/2 POUNDS

Step-by-step explanation:

3/8 - LIGHTEST SQUASH

7/8 - HEAVIEST SQAUSH

SUBTRACT THEM YOU GET 4/8

4/8 SIMPLIFIED IS 1/2

THUS THE SQUASH ARE 1/2 POUND DIFERENCE

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Alex_Xolod [135]

Answer:

18.3824%

Step-by-step explanation:

Sorry, this may be wrong. I am using an old skill, so it may be wrong.

3 0
2 years ago
(Minimum or maximum), <br> (4, 8, 7, 23.)
lisov135 [29]
8 minimum wage for the year to come round hea again and the
8 0
3 years ago
Read 2 more answers
Please I need the correct answer please
Rzqust [24]

(<u>−1</u>

2  )(n^3)+

<u>1</u>

2 n^2+4.6n+(−

<u>1</u>

2)(n^3)+

<u>1</u>

2  n^2+4.5n

=

<u>−1</u>

2  n^3+

1

2  n^2+4.6n+

−1

2  n^3+

1

2  n^2+4.5n

Combine Like Terms:

=

<u>−1</u>

2  n^3+

<u>1</u>

2  n^2+4.6n+

<u>−1</u>

2  n^3+

<u>1</u>

2  n^2+4.5n

=(<u>−1</u>

  2   n^3+

<u>−1</u>

2   n^3)+(

<u>1</u>

2  n^2+

<u>1</u>

2   n^2)+(4.6n+4.5n)

=−n^3+n^2+9.1n

Answer:

=−n^3+n^2+9.1n

     Everything underlined means its a  fraction/divided hope this helps <em>:D</em>

8 0
3 years ago
How would you go about finding the solution to this system of three equations in three variables? Be specific. For example, you
kupik [55]
I:2x – y + z = 7
II:x + 2y – 5z = -1
III:x – y = 6

you can first use III and substitute x or y to eliminate it in I and II (in this case x):
III: x=6+y
-> substitute x in I and II:
I': 2*(6+y)-y+z=7
12+2y-y+z=7
y+z=-5

II':(6+y)+2y-5z=-1
3y+6-5z=-1
3y-5z=-7

then you can subtract II' from 3*I' to eliminate y:
3*I'=3y+3z=-15

3*I'-II':
3y+3z-(3y-5z)=-15-(-7)
8z=-8
z=-1

insert z in II' to calculate y:
3y-5z=-7
3y+5=-7
3y=-12
y=-4

insert y into III to calculate x:
x-(-4)=6
x+4=6
x=2

so the solution is
x=2
y=-4
z=-1
3 0
3 years ago
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Novay_Z [31]

Answer:

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