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o-na [289]
3 years ago
5

Prove the following statement (1-sinA)×sin²A/(1+cosA)×cos²A=1-cosA/1+sinA​

Mathematics
1 answer:
shepuryov [24]3 years ago
4 0

Answer:

<u>(1-sinA) * Sin²A</u>

(1+cosA) * Cos²A

<u>(1-sinA) * (1-Cos²A)</u>

(1+cosA) * (1-Sin²A)

<u>(1-sinA) * (1-CosA)(1+CosA)</u>

(1+cosA) * (1-SinA) (1+SinA)

<u>(1-CosA)</u>

(1+SinA)

proved

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I am Lyosha [343]

Answer:

f= -.032

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3 years ago
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The roots of x2 -( ) +34 are 5 -/+3i
asambeis [7]

Answer: 10

Step-by-step explanation:

Vieta's formulas :

5 - 3i + 5 + 3i = b ⇒ b =10

Now, we have the equation:

x² -10x + 34 = 0

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7 0
3 years ago
Please help explanation if possible
Nookie1986 [14]

Answer:

y=-3x+5

Step-by-step explanation:

concepts: y=mx+b is slope intercept equation formula

m=slope

b= y intercept

Therefore we need to find the slope and y intercept

first find slope

m=\frac{y2-y1}{x2-x1}

let y2 be 2

let y1 be -4

let x2 be 1

let x1 be 3

\frac{2-(--4)}{1-3}

6/-2 = -3

m=-3

slope is -3

y= -3x+ b

we need find y intercept now

just plug in (1,2) into that equation

2=-3(1)+b

b=5

y=-3x+5

5 0
3 years ago
Write the equation of the circle centered at (3, 5) and tangent to x = -1. I need help pleasse
Annette [7]

Answer:

(x - 3)² + (y - 5)² = 16

Step-by-step explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

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given (h, k ) = (3, 5 ) we require to find r

r is the distance from the centre to a point on the circle

given x = - 1 is a tangent the r is the distance between the x- coordinates

r = | 3 - (- 1) | = | 3 + 1 | = | 4 | = 4

then equation of circle is

(x - 3)² + (y - 5)² = 4² , that is

(x - 3)² + (y - 5)² = 16

7 0
2 years ago
Which point is located at (4, -2)?
larisa86 [58]
Point B is located at point (4, -2)
7 0
2 years ago
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