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Law Incorporation [45]
3 years ago
14

If the point (3, 10) is on f(x), which point will be on f(x)+3? please help me :(

Mathematics
1 answer:
Vinil7 [7]3 years ago
5 0

Answer:

(3,13)

Step-by-step explanation:

so, you would add 3 to the y value

the point would be (3,13)

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Step-by-step explanation:

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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 > 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)

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3 years ago
X2 + 4x - 9 = 5x + 3<br> O A. 4<br> O B. -3<br> O C. -7<br> O D. 5<br> E. -2<br> F-4
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Answer:

{x}^{2}  + 4x - 9 = 5x + 3 \\  {x }^{2}  - x - 12 \\  {x}^{2}  - 4x + 3x - 12 \\ x(x - 4) + 3(x - 4) \\ (x - 4)(x + 3) \\ x = 4 \: and - 3

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A DNA molecule that is produced by combining DNA from different sources or organisms is called A) marker DNA. B) polymerase DNA.
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It is C I just studied DNA. Recombinant DNA is the combination of DNA  from 2 different sources. We have Recombinant DNA because we all have jeans(DNA) from both of our parents. Chemical Reaction "hormones".
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6 0
3 years ago
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Correct Answer will get BRAINLIEST!
frozen [14]

Answer:

The degree measure of ∠ACP is 112.5°

Step-by-step explanation:

The total area of the given diagram is the sum of two semicircles with arc AB and arc CB having radius R and r respectively

Where:

R = AC

r = DB

R = 2 × r

Therefore the area of the semicircles are given as follows;

For the semicircle with arc AB, we have;

Area, A₁ = π × R²/2 = π × (2×r)²/2 = 2×π×r²

For the semicircle with arc CB, we have;

Area, A₂ = π × r²/2 =  1/2×π×r²

The ratio of the two semicircles is presented in the following relation;

\dfrac{A_1}{A_2} = \dfrac{2 \cdot \pi \cdot r^2}{\dfrac{1}{2}  \cdot \pi \cdot r^2} = \dfrac{2}{\dfrac{1}{2} } = 2 \times \dfrac{2}{1}  = 4

Therefore, the area of A₁ is four times that of A₂ or A₁ =  4 × A₂

The total area of the given diagram = A₁ + A₂ = 4 × A₂ + A₂ = 5·A₂

∴ Half of the area of the diagram, A_H = 5·A₂/2 = 2.5·A₂ = 2.5 × 1/2×π×r² = 1.25×π×r²

The ratio of half of the diagram of the figure to the area of the semicircle with arc AB is found as follows;

A_H/A₁ = (1.25×π×r²)/(2×π×r²) = 5/8

Therefore, the half of the diagram of the figure given by segment PAC is equivalent to 5/8 of the semicircle with arc AB

Given that the arc AB subtends an angle of 180° at the center (angle subtended by a semicircle), the arc AP will subtend 5/8×180 = 112.5°

To verify we have;

Area of a segment of a circle is presented in the following relation;

\dfrac{\theta}{360} \times \pi  \times r^2

As segment PAC is 5/8 of a semicircle, it is therefore 5/(8×2) or 5/16 of the whole circle

Hence;

\dfrac{5}{16} \times \pi \times r^2 =  \dfrac{\theta}{360} \times \pi  \times r^2

\dfrac{5}{16} =  \dfrac{\theta}{360}

\theta \dfrac{}{} =  \dfrac{360 \times 5}{16} = 112.5 ^ {\circ}

Therefore the degree measure of ∠ACP is 112.5°.

5 0
3 years ago
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