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3 years ago

PLEASE ANSWER THIS I DIDN'T GET ANYTHING DUE IN 2 HOURS!!!!!

Mathematics

2 answers:

NikAS [45]3 years ago

5 0

Answer:

Step-by-step explanation:

√(1 - cos^2x) = sin x

and

√(1 - sin^2x) = cos x

So the expression becomes:

sin x / sin x + cos x / cos x

= 2

Aliun [14]3 years ago

3 0

Answer:

Step-by-step explanation:

using the trigonometric identity

• sin²x + cos²x = 1

⇒ sinx = √(1 - cos²x) and cosx = √(1 - sin²x)

We can express as

$\frac{sinx}{sinx}$ + cosx / cosx = 1 + 1 = 2 → H

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Vesnalui [34]

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Solve the equation : 2x + 7 = 21 and find the value of x. Is 2x + 7 = 21 a first degree equation or not?

Paraphin [41]

Answer:

The proposed equation is solved as follows:

2X + 7 = 21

2X = 21 - 7

2X = 14

X = 14 / 2

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The value of X is 7.

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2 years ago

From first principles, find the indicated derivatives

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By definition of the derivative,

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{(s + h)^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{s^3+3s^2h+3sh^2+h^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}$

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2 years ago

Greatest common factor of 6,12, and 21

kiruha [24]

Answer:

Step-by-step explanation:

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2 years ago

Find the orthocenter for the triangle described by each set of vertices.

givi [52]

Answer:

The orthocentre of the given vertices ( 2 , -3.5)

Step-by-step explanation:

Step(i):-

The orthocentre is the intersecting point for all the altitudes of the triangle.

The point where the altitudes of a triangle meet is known as the orthocentre.

Given Points are K (3.-3), L (2,1), M (4,-3)

The Altitudes are perpendicular line from one side of the triangle to the opposite vertex

The altitudes are MN , KO , LP

step(ii):-

Slope of the line

$KL = \frac{y_{2}-y_{1} }{x_{2}-x_{1} } = \frac{1-(-3)}{2-3} = -4$

The slope of MN =

The perpendicular slope of KL

= $\frac{-1}{m} = \frac{-1}{-4} = \frac{1}{4}$

The equation of the altitude

$y - y_{1} = m( x-x_{1} )$

$y - (-3) = \frac{1}{4} ( x-4 )$

4y +12 = x -4

x - 4 y -16 = 0 ...(i)

Step(iii):-

Slope of the line

$LM = \frac{y_{2}-y_{1} }{x_{2}-x_{1} } = \frac{-3-1}{4-2} = -2$

The slope of KO =

The perpendicular slope of LM

= $\frac{-1}{m} = \frac{-1}{-2} = \frac{1}{2}$

The equation of the altitude

$y - y_{1} = m( x-x_{1} )$

The equation of the line passing through the point K ( 3,-3) and slope

m = 1/2

$y - (-3) = \frac{1}{2} ( x-3 )$

2y +6 = x -3

x - 2y -9 =0 ....(ii)

Solving equation (i) and (ii) , we get

subtracting equation (i) and (ii) , we get

x - 4y -16 -( x-2y-9) =0

- 2y -7 =0

-2y = 7

y = - 3.5

Substitute y = -3.5 in equation x -4y-16=0

x - 4( -3.5) - 16 =0

x +14-16 =0

x -2 =0

x = 2

The orthocentre of the given vertices ( 2 , -3.5)

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3 years ago