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scoray [572]
3 years ago
8

Solve for t. d=3+at^2

Mathematics
1 answer:
GalinKa [24]3 years ago
8 0

Answer:

it's the 2nd one

Step-by-step explanation:

d = 3 + at^2

-3 | -3

d - 3 = at^2

d - 3/a = t^2

√d - 3/a = √t^2

t = √d - 3/a

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The Cartesian coordinates of a point are given. (a) (−5, 5) (i) Find polar coordinates (r, θ) of the point, where r &gt; 0 and 0
Alex73 [517]

Answer:

a)

(i) The coordinates of the point in polar form is (5√2 , 7π/4)

(ii) The coordinates of the point in polar form is (-5√2 , 3π/4)

b)

(i) The coordinates of the point in polar form is (6 , π/3)

(ii) The coordinates of the point in polar form is (-6 , 4π/3)

Step-by-step explanation:

* Lets study the meaning of polar form

- To convert from Cartesian Coordinates (x,y) to Polar Coordinates (r,θ):

1. r = √( x2 + y2 )

2. θ = tan^-1 (y/x)

- In Cartesian coordinates there is exactly one set of coordinates for any

 given point

- In polar coordinates there is literally an infinite number of coordinates

 for a given point

- Example:

- The following four points are all coordinates for the same point.

# (5 , π/3) ⇒ 1st quadrant

# (5 , −5π/3) ⇒ 4th quadrant

# (−5 , 4π/3) ⇒ 3rd quadrant

# (−5 , −2π/3) ⇒ 2nd quadrant

- So we can find the points in polar form by using these rules:

 [r , θ + 2πn] , [−r , θ + (2n + 1) π] , where n is any integer

 (more than 1 turn)

* Lets solve the problem

(a)

∵ The point in the Cartesian plane is (-5 , 5)

∵ r = √x² + y²

∴ r = √[(5)² + (-5)²] = √[25 + 25] = √50 = ±5√2

∵ Ф = tan^-1 (y/x)

∴ Ф = tan^-1 (5/-5) = tan^-1 (-1)

- Tan is negative in the second and fourth quadrant

∵ 0 ≤ Ф < 2π

∴ Ф = 2π - tan^-1(1) ⇒ in fourth quadrant r > 0

∴ Ф = 2π - π/4 = 7π/4

OR

∴ Ф = π - tan^-1(1) ⇒ in second quadrant r < 0

∴ Ф = π - π/4 = 3π/4

(i) ∵ r > 0

∴ r = 5√2

∴ Ф = 7π/4 ⇒ 4th quadrant

∴ The coordinates of the point in polar form is (5√2 , 7π/4)

(ii) r < 0

∴ r = -5√2

∵ Ф = 3π/4 ⇒ 2nd quadrant

∴ The coordinates of the point in polar form is (-5√2 , 3π/4)

(b)

∵ The point in the Cartesian plane is (3 , 3√3)

∵ r = √x² + y²

∴ r = √[(3)² + (3√3)²] = √[9 + 27] = √36 = ±6

∵ Ф = tan^-1 (y/x)

∴ Ф = tan^-1 (3√3/3) = tan^-1 (√3)

- Tan is positive in the first and third quadrant

∵ 0 ≤ Ф < 2π

∴ Ф = tan^-1 (√3) ⇒ in first quadrant r > 0

∴ Ф = π/3

OR

∴ Ф = π + tan^-1 (√3) ⇒ in third quadrant r < 0

∴ Ф = π + π/3 = 4π/3

(i) ∵ r > 0

∴ r = 6

∴ Ф = π/3 ⇒ 1st quadrant

∴ The coordinates of the point in polar form is (6 , π/3)

(ii) r < 0

∴ r = -6

∵ Ф = 4π/3 ⇒ 3rd quadrant

∴ The coordinates of the point in polar form is (-6 , 4π/3)

6 0
3 years ago
How do I solve -0.4r=16?
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Answer:

r = -40

Step-by-step explanation:

1. Move constants to the other side of the equation

r = 16/-.04

2. Simplify (by calculator)

You can use a calculator and plug in 16/-.04

r = -40

2. Simplify (by fraction)

You can turn -.4 into a fraction.

-.04 = - 4/10

r = 16/-(4/10)

r = 16*10/-4

r = -40

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3 years ago
HOW DO YOU DO THIS? PLEASE HELP
Maurinko [17]
What do you need help with there is no picture
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3 years ago
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