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NARA [144]
4 years ago
5

How do you find the slope of a point

Mathematics
2 answers:
torisob [31]4 years ago
8 0
Two points You mean? if So it's Rise/Run. E.G; Y2-Y1/X2-X1
Hoochie [10]4 years ago
5 0
A point doesn't have slope. Once you have TWO points, then the line between them has a slope. The slope is (how far the line rises from one point to the other) divided by (the level distance between them).
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Whats answere y+2=-3 (x - 4)​
faust18 [17]

Answer:

 Slope = 6.000/2.000 = 3.000

 x-intercept = 10/3 = 3.33333

 y-intercept = -10/1 = -10.00000

Step-by-step explanation:

Slope is defined as the change in y divided by the change in x. We note that for x=0, the value of y is -10.000 and for x=2.000, the value of y is -4.000. So, for a change of 2.000 in x (The change in x is sometimes referred to as "RUN") we get a change of -4.000 - (-10.000) = 6.000 in y. (The change in y is sometimes referred to as "RISE" and the Slope is m = RISE / RUN)

4 0
3 years ago
Read 2 more answers
5 puonds of grapes cost 9.35 how much would 2 pounds cost
Deffense [45]

Answer:

$3.74

Step-by-step explanation:

5 pounds is 9.35 means 1 pound is 1.87 dollars

2 pounds is 3.74 dollars

5 0
3 years ago
The expression sin 57° is equal to
ValentinkaMS [17]

Answer:

cos(33^{\circ})

Step-by-step explanation:

Given

sin(57^{\circ})

Required

Determine an equivalent expression

In trigonometry:

sin(\theta)= cos(90^{\circ} - \theta)

In sin(57^{\circ})

\theta=57^{\circ}

Substitute 57^{\circ} for \theta

in sin(\theta)= cos(90^{\circ} - \theta)

sin(57^{\circ})= cos(90^{\circ} - 57^{\circ})

sin(57^{\circ})= cos(33^{\circ})

Hence, the equivalent expression is: cos(33^{\circ})

5 0
4 years ago
Please help me with this math
natulia [17]

The slope is -1/2 or -0.5

3 0
2 years ago
53% of what number is 384? <br> i got 724.52 how do i round it to the hundredths place?
aleksandr82 [10.1K]

Answer:

thats right.

Step-by-step explanation:

the first place after the decimal is tenths, second, hundredths, third thousands, and so on

6 0
3 years ago
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