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Gemiola [76]
3 years ago
11

What should be added to 8 to make the sum 0

Mathematics
2 answers:
Jlenok [28]3 years ago
7 0
Y+8=0

well, y can't be a positive number.
think:
does 8-8=0?
yes, it does.
so you would do:

x+y=z

X=8
y= -8
z=0

(sorry if I confused you!)
Mars2501 [29]3 years ago
3 0
-8 because when you add a negative number your basically subtracting its absolute value.
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wlad13 [49]
The correct answer for this problem is D.
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Paul has 2/3 as many postcards as Shawn. The number of postcards Shawn has is 4/5 of the number of postcards Tim has. If the thr
ahrayia [7]

Tim has 56 more postcards than Paul.

Step-by-step explanation:

Total postcards boys have = 280

Let,

x = Paul's postcards

y = Shawn's postcards

z = Tim's postcards

According to given statement;

x + y + z = 280      Eqn 1

x=\frac{2}{3}y                   Eqn 2

y = \frac{4}{5}z                   Eqn 3

Putting value of y from Eqn 3 in Eqn 2

x=\frac{2}{3}*\frac{4}{5}z\\\\x=\frac{8}{15}z            Eqn 4

Putting value of x and y from Eqn 4 and Eqn 3 in Eqn 1

\frac{8}{15}z+\frac{4}{5}z+z=280\\\\\frac{8z+12z+15z}{15}=280\\\\\frac{35z}{15}=280\\\\

Multiplying both sides by \frac{15}{35}

\frac{15}{35}*\frac{35}{15}z=280*\frac{15}{35}\\z=120

Putting z=120 in Eqn 3

y=\frac{4}{5}*120\\y=96

Putting y=96 in Eqn 2

x=\frac{2}{3}*96\\x=64

Difference of Tim and Paul = 120-64 = 56

Tim has 56 more postcards than Paul.

Keywords: linear equation, substitution method

Learn more about substitution method at:

  • brainly.com/question/10557938
  • brainly.com/question/10600222

#LearnwithBrainly

7 0
3 years ago
I need help with number 1
Sophie [7]
Changing those words into an equation, we get:
7+3x = 19
Subtract both sides by 7
3x = 12
Divide both sides by 3
x = 4
That number is 4. Have a nice day! :)
6 0
3 years ago
The area, a, of a square whose side length is n
svetlana [45]

Answer:

a*n=area

Step-by-step explanation:

3 0
3 years ago
How can 38/6 be expressed as a decimal?
seraphim [82]
38/6 = 19/3
divide 19 by 3 :-

= 6.333....

The 3 is recurring
8 0
3 years ago
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