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egoroff_w [7]
3 years ago
11

Rewrite the expression as an equivalent expression that does not contain powers of trigonometric functions greater than 1.

Mathematics
1 answer:
Gnesinka [82]3 years ago
6 0
Cos^4(x) = 
(cos²x)² = 
(1-sen²x)² = 
1² - 2*1*sen²x + sen^4x = 
1 - 2sen²x + sen^4x
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11-x+3y=8 solve for y
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Isolate the variable by dividing each side by factors that don't contain the variable.

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The table represents the observed number of leaves y on x branches of a tree. Calculate the rate of change between 3 and 4 branc
Dennis_Churaev [7]

Answer:

<h2>25</h2>

Step-by-step explanation:

The rate of change of number of tree y on x branches of a tree is expressed as shown;

m = Δy/Δx where;

m is the rate of change

Δy is the change in number of leaves  between 3 and 4

Δx is change in number of branches  between 3 and 4

m = Δy/Δx = y₂-y₁/x₂-x₁

From table between 3 and 4, y₂ = 100, y₁ = 75, x₂ = 4, x₁ = 3

m = 100-75/4-3

m = 25/1

m = 25

<em>Hence the rate of change between 3 and 4 branches is 25</em>

5 0
3 years ago
Two points in the Cartesian plane are A(2.00 m, −4.00 m) and B(−3.00 m, 3.00 m). Find the distance between them and their polar
Brrunno [24]

The distance between the two points is d=8.6m

The polar coordinate of A is \left(4.47,296.57\right)

The polar coordinate of B is \left(4.24,135\right)

Explanation:

The two points are A(2,-4) and B(-3,3)

The distance between two points is given by,

d=\sqrt{(2+3)^{2}+(-4-3)^{2}}\\d=\sqrt{(5)^{2}+(-7)^{2}}\\d=\sqrt{25+49}\\d=8.6

Thus, the distance between the two points is d=8.6m

The polar coordinates of A can be written as (Distance, tan^{-1} \frac{y}{x} )

Distance = \sqrt{x^{2} +y^{2} }

Substituting A(2,-4), we get,

Distance = \sqrt{2^{2} +(-4)^{2} }=\sqrt{4+16 }=4.47

tan^{-1} \frac{y}{x} =tan^{-1} \frac{-4}{2}=-63.43

To make the angle positive, let us add 360,

\theta=360-63.43=296.57

The polar coordinate of A is \left(4.47,296.57\right)

Similarly, The polar coordinate of B can be written as (Distance, tan^{-1} \frac{y}{x} )

Distance = \sqrt{x^{2} +y^{2} }

Substituting B(-3,3), we get,

Distance = \sqrt{(3)^{2}+(3)^{2}}=4.24

tan^{-1} \frac{y}{x} =tan^{-1} \frac{3}{-3}=-45

To make the angle positive, let us add 360,

\theta=180-45=135^{\circ}

The polar coordinate of B is \left(4.24,135\right)

5 0
3 years ago
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