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alekssr [168]
3 years ago
11

0.6(y - 0.2) = 3 - 0.2(y - 1)

Mathematics
2 answers:
otez555 [7]3 years ago
8 0

Answer:

y=4.15

Step-by-step explanation:

0.6(y - 0.2) = 3 - 0.2(y - 1)

0.6y - 0.12 = 3 - 0.2y +0.2

        +0.12 =               +0.12

0.6y =3 - 0.2y +0.32

+0.2y = + 0.2y

0.8y/0.8 = 3.32/0.8

y= 4.15

inna [77]3 years ago
7 0

Answer: y= 4.15

You have to isolate the variable by itself

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There are 946 milliliters in a quart. There are 2 pints in a quart. How many milliliters are in a pint?​
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Which of the following could be points on the unit circle: A. (4/3, 4/5) B. (5/13, 12/13)
IRINA_888 [86]

Answer:

Option (2) and (4) are correct.

(\frac{5}{13},\frac{12}{13} ) and  (\frac{6}{7},\frac{\sqrt{13}}{7} ) are points on the unit circle.

Step-by-step explanation:

Given : Some points of circles.

We have to choose which points could be points on the unit circle.

We know, the equation of circle is

x^2+y^2=r^2  ..........(1)

where r is radius of circle.

We check each point for x and y values on the equation of circle and see which point gives radius = 1

Thus,

1) (\frac{4}{3},\frac{4}{5} )

Put  in LHS of  (1) , we have,

(\frac{4}{3})^2+(\frac{4}{5})^2

Simplify, we have,

=\frac{16}{9}+\frac{16}{25}

=\frac{544}{225}\neq 1

Thus,  (\frac{4}{3},\frac{4}{5} ) is not a point on the unit circle.

2) (\frac{5}{13},\frac{12}{13} )

Put  in LHS of  (1) , we have,

(\frac{5}{13})^2+(\frac{12}{13})^2

Simplify, we have,

=\frac{25}{169}+\frac{144}{169}

=\frac{169}{169}= 1

Thus, (\frac{5}{13},\frac{12}{13} ) is a point on the unit circle.

3) (\frac{1}{3},\frac{2}{3} )

Put  in LHS of  (1) , we have,

(\frac{1}{3})^2+(\frac{2}{3})^2

Simplify, we have,

=\frac{1}{9}+\frac{4}{9}

=\frac{5}{9}\neq 1

Thus,  (\frac{1}{3},\frac{2}{3} ) is not a point on the unit circle.

4)  (\frac{6}{7},\frac{\sqrt{13}}{7} )

Put  in LHS of  (1) , we have,

(\frac{6}{7})^2+(\frac{\sqrt{13}}{7})^2

Simplify, we have,

=\frac{36}{49}+\frac{13}{49}

=\frac{49}{49}= 1

Thus,  (\frac{6}{7},\frac{\sqrt{13}}{7} ) is a point on the unit circle.

Thus, (\frac{5}{13},\frac{12}{13} ) and  (\frac{6}{7},\frac{\sqrt{13}}{7} ) are points on the unit circle.

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3 years ago
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Answer:

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Step-by-step explanation:

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