What is the point used in the equation of the line y+4=1/2(x-2)

Mathematics

1 answer:

Ket [755]3 years ago

8 0

Answer:

What is the point used in the equation of the line y+4=1/2(x-2)

The other format for straight-line equations is called the "point-slope" form. For this one, they give you a point (x1, y1) and a slope m, and have you plug it into this formula:

y - y1 = m(x - x1)

Don't let the subscripts scare you. They are just intended to indicate the point they give you. You have the generic "x" and generic "y" that are always in your equation, and then you have the specific x and y from the point they gave you; the specific x and y are what is subscripted in the formula. Here's how you use the point-slope formula

They've given me m = 4, x1 = -1, and y1 = -6. I'll plug these values into the point-slope form, and solve for "y=":

y - y1 = m(x - x1)

y - (-6) = (4)(x - (-1))

y + 6 = 4(x + 1)

y + 6 = 4x + 4

y = 4x + 4 - 6

y = 4x - 2

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See question in attached photo.<br>Answer question 4b and 5c

sdas [7]

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Answer:

4b: 8.6 m/s²; 1.3×10^5 N; 2.1×10^4 N decrease

5c: -1.6×10^12 J; -1.6×10^12 J

Step-by-step explanation:

i) The acceleration due to gravity is inversely proportional to the square of the distance between the objects. The distance to the shuttle is ...

1 + (5×10^5)/(6.4×10^5) = 69/64 . . . times the radius of the earth

Then the acceleration due to gravity at the height of the space shuttle is about ...

(10 m/s²)(64/69)² ≈ 8.6 m/s²

ii) The weight of the space shuttle at that height is about ...

F = ma = (15000 kg)(8.6 m/s²) ≈ 1.3×10^5 N

iii) The loss of weight will be ...

ΔF = m(a1 -a0) = (15000 kg)(10 m/s² -8.6 m/s²)

= 1.5×10^4×1.4 N = 2.1×10^4 N

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i) The gravitational potential energy is given by ...

U = -GMm/r

where M and m are the mass of the earth and the rocket, respectively.

U = -(6.67×10^-11)(6.0×10^24)(2.5×10^4)/(6.4×10^6) ≈ -1.6×10^12 J

ii) At a height of 3×10^4 m, the denominator in the above expression changes from 6.4×10^6 to 6.43×10^6. This changes the gravitational potential energy by a factor of 6.4/6.43 to -1.6×10^12 J

(Note: we're carrying only 2 significant figures in the result in accordance with the rules for precision in such calculations. The change is noticeable at the level of the 4th significant figure, less than 1/2%.)

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2 years ago

A bus covers a distance of 2.45 km in 2 min. What is the speed of bus?

Vladimir [108]

Answer:

The speed of the bus is 74.242 kilometers per hour.

Step-by-step explanation:

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