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lawyer [7]
2 years ago
15

By , cos(θ) = StartFraction x Over r EndFraction, and by , sin(θ) = StartFraction y Over r EndFraction. Multiplying both sides o

f the above equations by r, we get that x = r cos(θ) and y = r sin(θ). The states that x2 + y2 = r2. By , we have [r cos(θ)]2 + [r sin(θ)]2 = r2. Applying the , the equation can be written as r2 cos2(θ) + r2 sin2(θ) = r2. Dividing both sides of the equation by r2 results in cos2(θ) + sin2(θ) = 1.
Mathematics
1 answer:
Marianna [84]2 years ago
8 0

Answer:

definition of cosine, definition of sine, pythagorean theorem, substitution, laws of exponents.

Step-by-step explanation:

You might be interested in
Manuela solved the equation 3−2|0.5x+1.5|=2 for one solution. Her work is shown below. 3−2|0.5x+1.5|=2 −2|0.5x+1.5|=−1 |0.5x+1.5
lions [1.4K]

Answer:

Step-by-step explanation:

We'll just work on solving both so you can see what's involved in solving an absolute value equation. Because an absolute value is a distance, we can have that distance being both to the right on the number line of the number in question or to the left. For example, from 2 on the number line, the numbers that are 5 units away are 7 and -3. Using that logic, we will simplify the equation down so we can set up the 2 basic equations needed to solve for x.

If  3-2|.5x+1.5|=2 then

-2|.5x+1.5|=-1  What you need to remember here is that you cannot distribute into a set of absolute values like you would a set of parenthesis. The -2 needs to be divided away:

|.5x+1.5|=.5

Now we can set up the 2 main equations for this which are

.5x + 1.5 = .5  and .5x + 1.5 = -.5

Knowing that an absolute value will never equal a negative number (because absolute values are distances and distances will NEVER be negative), once we remove the absolute value signs we can in fact state that the expression on the left can be equal to a negative number on the right, like in the second equation above.

Solving the first one:

.5x = -1 so

x = -2

Solving the second one:

.5x = -2 so

x = -4

7 0
2 years ago
Read 2 more answers
What is the ratio of the area of sector ABC to the area of sector DBE?
shusha [124]

We have to find the" ratio of the area of sector ABC to the area of sector DBE".

Now,

the general formula for the area of sector is

Area of sector= 1/2 r²θ

where r is the radius and θ is the central angle in radian.


180°= π rad

1° = π/180 rad


For sector ABC, area= 1/2 (2r)²(β°)

= 1/2 *4r²*(π/180 β)

= 2r²(π/180 β)

For sector DBE, area= 1/2 (r)²(3β°)

= 1/2 *r²*3(π/180 β)

= 3/2 r²(π/180 β)

Now ratio,

Area of sector ABC/Area of sector DBE =\frac{2r^{2}*\ \frac{\pi}{180} beta}{3/2 r^{2}*\ \frac{\pi}{180}beta}

= 4/3

7 0
2 years ago
Read 2 more answers
Which is a stretch of an exponential growth function? f(x) = Two-thirds (two-thirds) Superscript x f(x) = Three-halves (two-thir
Arte-miy333 [17]

Answer:

f(x) = Three-halves (three-halves) Superscript x

f(x) = Two-thirds (three-halves) Superscript x

Step-by-step explanation:

Since, a function in the form of f(x) = ab^x

Where, a and b are any constant,

is called exponential function,

There are two types of exponential function,

  • Growth function : If b > 1,
  • Decay function : if 0 < b < 1,

Since, In

f(x) =\frac{2}{3}(\frac{2}{3})^x

\frac{2}{3} < 1

Thus, it is a decay function.

in f(x) =\frac{3}{2}(\frac{2}{3})^x

\frac{2}{3} < 1

Thus, it is a decay function.

in f(x) =\frac{3}{2}(\frac{3}{2})^x

\frac{3}{2} > 1

Thus, it is a growth function.

in f(x) =\frac{2}{3}(\frac{3}{2})^x

\frac{3}{2} > 1

Thus, it is a growth function.

5 0
2 years ago
Read 2 more answers
Given that a function, g, has a domain of -1 ≤ x ≤ 4 and a range of 0 ≤ g(x) ≤ 18 and that g(-1) = 2 and g(2) = 8, select the st
notka56 [123]

Options

  • (A)g(5) = 12
  • (B)g(1) = -2
  • (C)g(2) = 4
  • (D)g(3) = 18

Answer:

(D)g(3) = 18

Step-by-step explanation:

Given that the function, g, has a domain of -1 ≤ x ≤ 4 and a range of                       0 ≤ g(x) ≤ 18 and that g(-1) = 2 and g(2) = 8

Then the following properties must hold

  1. The value(s) of x must be between -1 and 4
  2. The values of g(x) must be between 0 and 18.
  3. g(-1)=2
  4. g(2)=9

We consider the options and state why they are true or otherwise.

<u>Option A: g(5)=12</u>

The value of x=5. This contradicts property 1 stated above. Therefore, it is not true.

<u>Option B: g(1) = -2 </u>

The value of g(x)=-2. This contradicts property 2 stated above. Therefore, it is not true.

<u>Option C: g(2) = 4 </u>

The value of g(2)=4. However by property 4 stated above, g(2)=9. Therefore, it is not true.

<u>Option D: g(3) = 18</u>

This statement can be true as its domain is in between -1 and 4 and its range is in between 0 and 18.

Therefore, Option D could be true.

3 0
2 years ago
Circle groups to show 3x(2x3)
docker41 [41]
2x3=6x3=18 u always do PEMDAS,
5 0
2 years ago
Read 2 more answers
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