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Kamila [148]
3 years ago
8

Which product shows the decimal placed correctly?

Mathematics
2 answers:
butalik [34]3 years ago
6 0
C)2.1 I at he correct answer
GaryK [48]3 years ago
5 0

Answer:

the answer is C __________________________________

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Please help me with this!
ivanzaharov [21]

1.  The first equation is - 2x + 5y = 0

Second equation is y = \frac{2}{5} x

5y = 2x

- 2x + 5y = 0

Hence, the two equations are equivalent.

2.  a_{1} = 2, a_{2} = - 2

b_{1} = -1,  b_{2} = -1

\frac{a_{1} }{a_{2}} =\frac{2}{-2} = -1

\frac{b_{1} }{b_{2}} = \frac{-1}{-1}  = 1

\frac{a_{1} }{a_{2}} \neq \frac{b_{1} }{b_{2}}

Hence, the equations are consistent.

3.   a_{1} = 4, a_{2} = 6

b_{1} = -1, b_{2} = -1

\frac{a_{1} }{a_{2}} =\frac{4}{6} = \frac{2 }{3}

\frac{b_{1} }{b_{2}} = \frac{-1}{-1} = 1

\frac{a_{1} }{a_{2}} \neq \frac{b_{1} }{b_{2}}

Hence, the equations are consistent.

4.  Equations can be re-arranged as:

x + y - 4 = 0 and

x + y + 6 = 0

a_{1} = 1, a_{2} = 1

b_{1} = 1, b_{2} = 1

c_{1} = -4, c_{2} = 6

\frac{a_{1} }{a_{2}} =\frac{1}{1} = 1

\frac{b_{1} }{b_{2}} =\frac{1}{1} = 1

\frac{c_{1} }{c_{2}} =\frac{-4}{6} = \frac{-2}{3}

\frac{a_{1} }{a_{2}} = \frac{b_{1} }{b_{2}} \neq \frac{c_{1} }{c_{2}}

Hence, the equations are inconsistent.

5.  If we multiply the first equation by 4, we will get,

2y = -4x + 20 which is the second equation.

Hence, the equations are equivalent.

4 0
4 years ago
The area of a regular hexagon is 15 cm2. What is the area of a similar hexagon with sides four times as large?
denis-greek [22]

Answer:

60 cm2

Step-by-step explanation:

15x4=60

8 0
3 years ago
Which expression can be used to find 80% of 120?
Leya [2.2K]
Answer
(0.8) (120)
80/100 is 80%
5 0
3 years ago
If there are 16 values in a data set in order from smallest to largest, what is the median of the data set?
satela [25.4K]
A.) The mean of the 8th and 9th values
4 0
3 years ago
What is the result of an increase by 2/3
velikii [3]

Answer:

y=1.6 repeatin

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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