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Zina [86]
3 years ago
6

Find the slope and y intercept of the line below Y-9x=1/2

Mathematics
1 answer:
ANTONII [103]3 years ago
7 0

Answer:

slope is 9 and the y-intercept is 1/2

Step-by-step explanation:

y - 9x = 1/2, then y = 9x + 1/2

therefore the slope is 9 and the y-intercept is 1/2

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Move vertex B to a different location. What is the sum of the angle measures of the quadrilateral?
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Which of the following options best describes the function graphed below?
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Nonlinear increasing
6 0
3 years ago
-9 more than a number results in -20
damaskus [11]

Answer:

-9+x=-20

Step-by-step explanation:

I am not so positive on this one, but if you are not doing inequalities, then this should be correct.

"More than" would insinuate that you are adding. In this class, you would have a number to replace that x, but since no number was given, you place the x in its place. x=-11, because when you "add" -9 with something to get -20, it would have to be -11. Negative numbers can be confusing because when you subtract a negative number by another negative, you end up with a negative. Like in this case. Normally, you would just put -11, so it would look like -9 - 11 = -20. But since this says "more than", unless you are doing inequalities, you add.

If you are doing inequalities, then your answer should be this:

-9 > x = -20

I hope this helps!

-No one

4 0
2 years ago
Two points in the Cartesian plane are A(2.00 m, −4.00 m) and B(−3.00 m, 3.00 m). Find the distance between them and their polar
Brrunno [24]

The distance between the two points is d=8.6m

The polar coordinate of A is \left(4.47,296.57\right)

The polar coordinate of B is \left(4.24,135\right)

Explanation:

The two points are A(2,-4) and B(-3,3)

The distance between two points is given by,

d=\sqrt{(2+3)^{2}+(-4-3)^{2}}\\d=\sqrt{(5)^{2}+(-7)^{2}}\\d=\sqrt{25+49}\\d=8.6

Thus, the distance between the two points is d=8.6m

The polar coordinates of A can be written as (Distance, tan^{-1} \frac{y}{x} )

Distance = \sqrt{x^{2} +y^{2} }

Substituting A(2,-4), we get,

Distance = \sqrt{2^{2} +(-4)^{2} }=\sqrt{4+16 }=4.47

tan^{-1} \frac{y}{x} =tan^{-1} \frac{-4}{2}=-63.43

To make the angle positive, let us add 360,

\theta=360-63.43=296.57

The polar coordinate of A is \left(4.47,296.57\right)

Similarly, The polar coordinate of B can be written as (Distance, tan^{-1} \frac{y}{x} )

Distance = \sqrt{x^{2} +y^{2} }

Substituting B(-3,3), we get,

Distance = \sqrt{(3)^{2}+(3)^{2}}=4.24

tan^{-1} \frac{y}{x} =tan^{-1} \frac{3}{-3}=-45

To make the angle positive, let us add 360,

\theta=180-45=135^{\circ}

The polar coordinate of B is \left(4.24,135\right)

5 0
3 years ago
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