Answer:
a.
<u>cos θ = 2/5, tan θ = ?</u>
- tan θ = sin θ / cos θ
- tan θ = √sin² θ / (2/5)
- tan θ= 5√(1 - cos²θ) / 2
- tan θ = 5√(1 - 4/25) / 2
- tan θ = 5√(21/25) / 2
- tan θ = √21 / 2
b.
<u>cosec θ = 7/3, cos θ = ?</u>
- cosec θ = 1/ sin θ
- cosec θ = 1/ √(1 - cos²θ)
- √(1 - cos²θ) = 1 / cosec θ
- 1 - cos²θ = (1 / (7/3))²
- cos²θ = 1 - 9 / 49
- cos²θ = 40/49
- cos θ = √40/49
- cos θ = 2√10/7
c.
<u>cot θ = 4/3, sec θ = ?</u>
- cot θ = cos θ / sin θ
- cos θ = cot θ * sin θ
- cos θ = 4/3 * √(1 - cos²θ)
- 9cos²θ = 16(1 - cos²θ)
- 25cos²θ = 16
- cos²θ = 16/25
- cos θ √16/25
- cos θ = 4/5
- sec θ = 1/ cos θ
- sec θ = 1/ (4/5)
- sec θ = 5/4
d.
<u>tan θ = 3, cosec θ = ?</u>
- sin²θ + cos²θ = 1
- 1 + cos²θ/sin²θ = 1/ sin²θ
- 1 + 1/tan²θ = cosec²θ
- 1 + 1/9 = cosec²θ
- cosec²θ = 10/9
- cosec θ = √(10/9)
- cosec θ = √10 / 3
Answer:
the greatest possible value for x is 6
Explanation:
Assume that the sides of the triangle are:
a = x
b = 3x
c = 20
We are given that c is the longest side.
For the triangle to be obtuse:
c^2 > a^2 + b^2
Substitute with the values of a, b and c and solve for x as follows:
c^2 > a^2 + b^2
(20)^2 > (x)^2 + (3x)^2
400 > x^2 + 9x^2
400 > 10 x^2
40 > x^2
Now, we will get the zeros of the function. This means that we will solve for x^2 = 40 as follows:
x^2 = 40
x = + or - √40
either x = 6.32
Or x = -6.32
Since we want x^2 < 40, therefore, the greatest possible value for x to satisfy this condition is 6
Hope this helps :)
it has no root
first make it common quadratic form
y = 5x² - 3x + 8
we take
a = 5
b = -3
c = 8
the to prove it doesn't have root. we using discriminant
D = b² - 4ac
D = (-3)² - 4*5*8
D = 9 - 160
D = - 151
D > 0 means the equation has 2 roots
D = 0 means the equation has 1 root
D < 0 do not have root because it doesn't have intercept on x-axis
It’s B.
Distribute the 7a and multiply it with the 3a and -4 a. With the exponents you add them.