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Black_prince [1.1K]
3 years ago
14

Graph this equation Y=3x

Mathematics
1 answer:
Zarrin [17]3 years ago
6 0
Your line on the graph should start at (0, 3). Then (1, 6), (2, 9), and so on.
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Which expression is equivalent to \sqrt(-80)?
DedPeter [7]
√-80 can be factored out, to see which factors can be square rooted. A pair of factors would be 16 and -5, so that would be 4√-5. But, negative numbers cannot be inside a square root. So, √-1 is the same thing as <em>i</em>, or <u>imaginary number</u>. So, the expression equivalent to √-80 is 4<em>i</em>√5.
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3 years ago
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3 years ago
The coordinate plane below represents a city. Points A through F are schools in the city. graph of coordinate plane. Point A is
Ray Of Light [21]
Part A;
There are many system of inequalities that can be created such that only contain points A and E in the overlapping shaded regions.

Any system of inequalities which is satisfied by (2, -3) and (3, 1) but is not satisfied by (-3, -4), (-4, 2), (2, 4) and (-2, 3) can serve.

An example of such system of equation is
y ≤ x
y ≥ -2x
The system of equation above represent all the points in the first quadrant of the coordinate system.
The area above the line y = -2x and to the right of the line y = x is shaded.



Part B:
It can be verified that points A and E are solutions to the system of inequalities above by substituting the coordinates of points A and E into the system of equations and see whether they are true.

Substituting A(2, -3) into the system we have:
-3 ≤ 2
-3 ≥ -2(2) ⇒ -3 ≥ -4
as can be seen the two inequalities above are true, hence point A is a solution to the set of inequalities.

Also, substituting E(3, 1) into the system we have:
1 ≤ 3
1 ≥ -2(3) ⇒ 1 ≥ -6
as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.



Part C:
Given that William can only attend a school in her designated zone and that William's zone is defined by y < −x - 1.

To identify the schools that William is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining William's zone.

For point A(2, -3): -3 < -(2) - 1 ⇒ -3 < -2 - 1 ⇒ -3 < -3 which is false

For point B(-3, -4): -4 < -(-3) - 1 ⇒ -4 < 3 - 1 ⇒ -4 < 2 which is true

For point C(-4, 2): 2 < -(-4) - 1 ⇒ 2 < 4 - 1 ⇒ 2 < 3 which is true

For point D(2, 4): 4 < -(2) - 1 ⇒ 4 < -2 - 1 ⇒ 4 < -3 which is false

For point E(3, 1): 1 < -(3) - 1 ⇒ 1 < -3 - 1 ⇒ 1 < -4 which is false

For point F(-2, 3): 3 < -(-2) - 1 ⇒ 3 < 2 - 1 ⇒ 3 < 1 which is false

Therefore, the schools that Natalie is allowed to attend are the schools at point B and C.
4 0
3 years ago
Find the area of the sector of a circle with radius 2 inches formed by a central angle of 4.3 radians: (in square inches)
aleksandr82 [10.1K]

Answer:

8.6 in²

Step-by-step explanation:

The area of the entire circle is A = πr², or A = π(2 in)².

This corresponds to a central angle of 2π radians, or 6.28 radians.

The area of the sector described above is just a certain fraction of that:

4.3 rad

-------------- * Area of circle = 0.685(4 * 3.14) = 8.6 in²

6.28 rad

3 0
3 years ago
The sum of the lengths of KLM is 16.14 cm. The length of KL is 6 cm and the length of LM is 4.9 cm. What is the length of KM?
maria [59]
C i believe USBDHSUAIS
8 0
2 years ago
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