Resolva os sistemas abaixo 2x+3y=80 3x+2y=70 X=8y X+4y=480

1 answer:

7 0

Solución 1: 2(x) + 3(y) = 80
solución 2: 3(x) + 2(y) = 70

3(y) + 2(x) = 80
2(y) + 3(x) = 70

2y = -3x + 70
y = -3x÷2 + 35

2x + 3(-3x÷ 2+35) = 80
-5x÷2 = -25
-5x = -50

5x = 50
x-intercept: 10
x = 10
y = -3x÷ 2+35

y = -(3÷2)(10)+35 = 20

x-interpcet: 10
y-intercept: 20
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
solución: x - 8y = 0
solución x + 4y = 480

-8y + x = 0
4y + x = 480

x-intercept (solución 2)
x = -4x + 480

x-intercept (solución 1)
4y + 180 - 8y = 0
-12y = 480

y-intercept: 40
x-intercept: 320

Buena Suerte!

You might be interested in

What is the simplified form of x+8/4-x+5/4

kap26 [50]

Answer:

Step-by-step explanation:

x+ 8/4 -x+ 5/4

= x+2+-x +5/4

Combine Like Terms:

x+2+-x+ 5/4

(x+x)+ ( 2 +5/4)

= 13/4

5 0

2 years ago

Read 2 more answers

Please help i really don't know the answer.

inn [45]

Answer:

95 degrees

Step-by-step explanation:

The degrees of a pentagon adds up to 540 so just subtract the angles you have from that to get the answer

8 0

3 years ago

Read 2 more answers

Inside a square with side length 10, two congruent equilateral triangles are drawn such that they share one side and each has on

Sphinxa [80]

Answer:

5 length

Step-by-step explanation:

The diagram attached shows two equilateral triangles ABC & CDE. Since both squares share one side of the square BDFH of length 10, then their lengths will be 5 each. To obtain the largest square inscribed inside the original square BDFH, it makes sense to draw two other equilateral triangles AGH & EFG at the upper part of BDFH with length equal to 5.

So, the largest square that can be inscribe in the space outside the two equilateral triangles ABC & CDE and within BDFH is the square ACEG.