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GREYUIT [131]
2 years ago
11

You have determined that the equations x+1=y​ecks plus one is equal to why and −3x+5=y​negative three ecks plus five is equal to

why form a system of equations that has one solution.
To determine the solution of this system, you can first What?
Mathematics
1 answer:
Anna [14]2 years ago
5 0

Answer:

b

Step-by-step explanation:

2022 on edge I just did it

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A square park has a diagonal walkway from one corner to another. If the walkway is 120 meters long, what is the approximate leng
Vlad1618 [11]

Answer:

  85 m

Step-by-step explanation:

The diagonal of a square is √2 times the length of the side. The park will have a side length of 120/√2 m ≈ 84.85 m, about 85 meters.

_____

The relations are ...

  diagonal = (√2)×(side length)

  side length = diagonal/√2 . . . . . . . . . divide the above equation by √2

5 0
3 years ago
A ring is now reduced to 840 saving 40% of orignal price what is orignal price
denis23 [38]
The multiplier for decreasing by 40% is 0.60. Since we want to know what was reduced by 40% (0.60), we divide the new price by the multiplier:

840 / 0.6 = 1,400

So the original price was 1,400
5 0
3 years ago
Which relation iS a function?
Gre4nikov [31]

Answer:

The one shaped like an upright U

(Top right corner)

Step-by-step explanation:

This would be a function because when you do the vertical line test, the inputs (x) are paired with one output (y)

Hope this Helps

5 0
2 years ago
Standform........ math
harina [27]
36.22
If I'm correct that's the answer 
8 0
3 years ago
Read 2 more answers
A) Show that the equation 20-x^3-7x^2=0 can be rearranged to give x=20/x^2 - 7
AVprozaik [17]

Answer:

Step-by-step explanation:

      20-x^3-7x^2=0

a)

  • Add x^3 to both sides of the equation:     20-7x^2=x^3
  • Divide both sides by x^2:    \frac{20}{x^2} -\frac{7x^2}{x^2} =\frac{x^3}{x^2}  ⇒  \frac{20}{x^2} -7 =x

b)

    x_1=\frac{20}{(x_0)^2} -7

⇒ x_1=\frac{20}{(-9)^2} -7

⇒ x_1=-\frac{547}{81}

    x_2=\frac{20}{(x_1)^2} -7

⇒ x_2=\frac{20}{(-\frac{547}{81})^2} -7

⇒ x_2=-6.561443673...

   x_3=\frac{20}{(x_2)^2} -7

⇒ x_3=\frac{20}{(-6.561443673...)^2} -7

⇒ x_3=-6.535451368...

c) approximation to the location of one of the roots of the equation.  Each iteration gives a slightly more accurate value of a root x.

4 0
2 years ago
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