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vlada-n [284]
3 years ago
12

Please help asap 25 pts

Mathematics
2 answers:
allsm [11]3 years ago
8 0

Answer:

There is no single answer to this question other than The first graph on the left is the answer. But you should read the explanation and memorize it.

Step-by-step explanation:

It's the first graph on the right. The points are always plotted (without exception) as (x value which means go along the x axis horizontally - left or right.), (y value which means to up (for a plus y) or down for a minus y. These facts are just memorized.

Summary

(x,y)

  • x goes either left or right.
  • x>0 goes right.  
  • x<0 goes left.
  • y goes up or down
  • y > 0 goes up
  • y < 0 goes down

Locations

First point (Please label this as first point)  - 3 (that goes left)  2 that goes up

You should be putting it in the upper left space.

Second Point. (-2,-2) that goes 2 to left and 2 down. You should put that point in the lower left space. There's only 1 point in the lower left space and that is this point.

Third Point (0,1) That's the only point on the y axis. It is above the x axis. It is all by itself on the y axis line. The other two points are done the same way. Please make sure you try them.


bekas [8.4K]3 years ago
7 0

{(-3, 2), (-2, -2), (0, 1), (2, 4), (4, -3)}

(x, y)

Answer: a.


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One of the legs of a right triangle measures 4 cm and its hypotenuse measures 18 cm. Find the measure of the other leg. If neces
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Answer:

√308 or 17.5

Step-by-step explanation:

c^2=a^2+b^2

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Substitute

18^2-4^2=b^2

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Step-by-step explanation:

1. You subtract 5-7 because of PEMDAS and you get -2

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3. Multiply 35x-2=-70

Parentheses/brackets

Exponents

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Svetlanka [38]

The x-intercepts are (0,0)

The y-intercepts are (0,0),(0,3),(0,-3)

Explanation:

The relation is x=y^{4} -9y^{2}

To find the x-intercept, let us substitute y=0 in the equation x=y^{4} -9y^{2}, we get,

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To find the y-intercept, let us substitute x=0 in the equation x=y^{4} -9y^{2}, we get,

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let us switch sides and solving, we get,

\begin{aligned}y^{4}-9 y^{2} &=0 \\y^{2}\left(y^{2}-9\right) &=0 \\y^{2} &=0, y^{2}-9=0\end{aligned}

Taking square root,

y=0 and \begin{aligned}y^{2}-9 &=0 \\y^{2} &=9 \\y &=\pm 3\end{aligned}

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Step-by-step explanation:

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