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3 years ago

What is the value of y in the following system? y=3x-5 6x+3y=15

Mathematics

2 answers:

charle [14.2K]3 years ago

8 0

Step-by-step explanation:

y=3x-5

6x+3y=15

6x+3(3x-5)=15

6x+9x-15=15

15x-15=15

+15 +15

15x=30

/15. /15

x=2

6(2)+3y=15

12+3y=15

-12. -12

3y=3

/3 /3

y=1

(2,1)

The answer is (2,1)

8090 [49]3 years ago

6 0

Step-by-step explanation:

6x + 3(3x - 5) = 15

6x + 9x - 15 = 15

15x - 15 = 15

+15 +15

15x = 30

/15 /15

x = 2

6(2) + 3y = 15

12 + 3y = 15

-12 -12

3y = 3

/3 /3

y = 1

(2, 1)

This should be right if it is please make me brainliest.

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3 years ago

Draw at least six different sized rectangles that have an area of 64 square units

olga_2 [115]

Let

x------> the length of the rectangle

y------> the width of the rectangle

we know that

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$A=x*y$

$A=64\ units^{2}$

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let's assume different values of x to get the different values of y

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see the draw in the attached figure N 1

case 2) For x=32 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/32$

$y=2\ units$

the dimensions of the rectangle are 32 units x 2 units

see the draw in the attached figure N 2

case 3) For x=16 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/16$

$y=4\ units$

the dimensions of the rectangle are 16 units x 4 units

see the draw in the attached figure N 3

case 4) For x=30 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/30$

$y=\frac{32}{15} =2\frac{2}{15} \ units$

the dimensions of the rectangle are 30 units x 2 (2/15) units

see the draw in the attached figure N 4

case 5) For x=40 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/40$

$y=1.60\ units$

the dimensions of the rectangle are 40 units x 1.60 units

see the draw in the attached figure N 5

case 6) For x=60 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/60$

$y=\frac{16}{15} =1\frac{1}{15} \ units$

the dimensions of the rectangle are 60 units x 1 (1/15) units

see the draw in the attached figure N 6

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3 years ago

Brainliest ,easy question

insens350 [35]

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3 years ago

What is the sum of the first 51 consecutive odd positive integers?

Angelina_Jolie [31]

We call:

$a_{n}$ as the set of the first 51 consecutive odd positive integers, so:

 $a_{n} = \{1, 3, 5, 7, 9...\}$

Where:
$a_{1} = 1$
$a_{2} = 3$
$a_{3} = 5$
$a_{4} = 7$
$a_{5} = 9$
and so on.

In mathematics, a sequence of numbers, such that the difference between two consecutive terms is constant, is called Arithmetic Progression, so:

3-1 = 2
5-3 = 2
7-5 = 2
9-7 = 2 and so on.

Then, the common difference is 2, thus:

 $a_{n} = \{ a_{1} , a_{1} + d, a_{1} + d + d,..., a_{1} + (n-2)d+d\}$

Then:

 $a_{n} = a_{1} + (n-1)d$

So, we need to find the sum of the members of the finite series, which is called arithmetic series:

There is a formula for arithmetic series, namely:

 $S_{k} = ( \frac{a_{1} + a_{k}}{2} ).k$

Therefore, we need to find:
 $a_{k} = a_{51}$

Given that $a_{1} = 1$ , then:

$a_{n} = a_{1} + (n-1)d = 1 + (n-1)(2) = 2n-1$

Thus:
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Lastly:

$S_{51} = ( \frac{1 + 101}{2} ).51 = 2601$

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