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posledela
3 years ago
13

Type the correct answer in each box. If necessary, use / for the fraction bar(s).

Mathematics
1 answer:
spayn [35]3 years ago
5 0

Answer:X=5/6 Y=8/5

Step-by-step explanation:

-6x +5y=3

+6x

5y=6x+3 (divide each number by 5)

Y=6/5x + 3/5 (Plug that into your x equation)

12x+15 (6/5x + 3/5) =34

(Multiply 15 by the numbers in the brackets you can break it down and do 15•6 divided by 5 )

15•6=90 divided by 5=18x

15•3= 45 divided by 5=9

(Add like variables)

12x+18x=30x

30x+9=34

(Subtract 9 from itself do the same to 34)

9-9= 0

34-9=25

Divide 30 by itself to get x by itself then divide 25 by 30

25 divided by 30= 5/6

X=5/6 (plug that into the y equation)

Y=6/5•5/6+3/5

Multiply 6/5 and 5/6 straight across because this is multiplication you don't need like denominators

6/5•5/6=30/30=1

Y=1+3/5

1/1 + 3/5

Here you need to have like denominators because it is addition so just use 5

5•1=5

5/5+3/5=8/5

Y=8/5

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What is the perimeter of the triangle shown on the coordinate plane, to the nearest tenth of a unit?
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Answer:

25.6 units

Step-by-step explanation: From the figure we can infer that our triangle has vertices A = (-5, 4), B = (1, 4), and C = (3, -4).

First thing we are doing is find the lengths of AB, BC, and AC using the distance formula:

d=\sqrt{(x_2-x_1)^{2} +(y_2-y_1)^{2}}

where

(x_1,y_1) are the coordinates of the first point

(x_2,y_2) are the coordinates of the second point

- For AB:

d=\sqrt{[1-(-5)]^{2}+(4-4)^2}

d=\sqrt{(1+5)^{2}+(0)^2}

d=\sqrt{(6)^{2}}

d=6

- For BC:

d=\sqrt{(3-1)^{2} +(-4-4)^{2}}

d=\sqrt{(2)^{2} +(-8)^{2}}

d=\sqrt{4+64}

d=\sqrt{68}

d=8.24

- For AC:

d=\sqrt{[3-(-5)]^{2} +(-4-4)^{2}}

d=\sqrt{(3+5)^{2} +(-8)^{2}}

d=\sqrt{(8)^{2} +64}

d=\sqrt{64+64}

d=\sqrt{128}

d=11.31

Next, now that we have our lengths, we can add them to find the perimeter of our triangle:

p=AB+BC+AC

p=6+8.24+11.31

p=25.55

p=25.6

We can conclude that the perimeter of the triangle shown in the figure is 25.6 units.

Read more on Brainly.com - brainly.com/question/12560433#readmore

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