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Lyrx [107]
3 years ago
8

Please help, thanks if you do :)

Mathematics
1 answer:
kati45 [8]3 years ago
8 0

Answer: (4,-2)

Step-by-step explanation: step1.- -1/2x=-x+2 (solving for x) x=4

Step 2- y=-1/2x4(substitute the value of X)

Step 3- y=-1/2x4 (solve the equation for y) =2

Step 4- (x,y)=(4,-2) simplify —>(x,y)=(4,-2)

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Obtain the general solution of<br> y ln x ln y dx + dy = 0
oksano4ka [1.4K]
Dy/dx = -y inx iny

(y in x in y) dx + dy = 0

(y in x in y ) dx = - dy

(in x )dx =  - (1/ y in y) dy

you just need to integrate both sides

Hope this helps
4 0
4 years ago
A seedless watermelon
MA_775_DIABLO [31]
$0.88

20%=0.2
0.2x4.4=0.88
4 0
3 years ago
How do you find the square root of a fraction? For example: what’s the square root of 115/100
Bingel [31]

Answer:

Take the square roots of the numerator and denominator separately.

Explanation:

y = \sqrt{\frac{115 }{100}}

y = \frac{\sqrt{115} }{10}

===============

If your instructor wants a decimal answer, take the square root of the numerator.

\frac{\sqrt{115} }{10} = \frac{10.72}{10}

\frac{\sqrt{115} }{10} = 1.072

4 0
3 years ago
Please help me please
Tamiku [17]

Answer:

(4,-3)

Step-by-step explanation:

is 4<3? no

is -2<3? yes but is 1>_ 8? no

etc etc

6 0
3 years ago
Drag each tile to the correct box.
wlad13 [49]

The correct match of each tile of the equation with its solution is:

  • n - 13 = - 12 →→→ 1
  • n/5 = -1/5  →→→ -1
  • n + 15 = - 10 →→→ -25

<h3>How do we match each tile to the correct box?</h3>

To match each tile to the correct box, we have to solve the arithmetic operations in the box, then drag the correct tile that matches our answer into the box.

From the image attached below;

1.

n - 13 = - 12

Let us add (+13) to both sides to eliminate (-13), i.e.

n - 13 + 13 = - 12 + 13

n = 1

2.

n/5 = -1/5  

multiply both side by 5

n/5 × (5) = -1/5 × (5)

n = -1

3.

n + 15 = - 10

n +15 - 15 = - 10 - 15

n = -25

Learn more about matching each tile to the correct box here:

brainly.com/question/17203448

#SPJ1

7 0
2 years ago
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