Let ABC be a triangle in the 3rd quadrant, right-angled at B.
So, AB-> Perpendicular BC -> Base AC -> Hypotenuse.
Given: sinθ=-3/5 cosecθ=-5/3
According to Pythagorean theorem, square of the hypotenuse is equal to the sum of square of the other two sides.
Therefore in triangle ABC, 〖AC〗^2=〖AB〗^2+〖BC〗^2 ------
--(1)
Since sinθ=Perpendicular/Hypotenuse ,
AC=5 and AB=3
Substituting these values in equation (1)
〖BC〗^2=〖AC〗^2-〖AB〗^2
〖BC〗^2=5^2-3^2
〖BC〗^2=25-9
〖BC〗^2=16
BC=4 units
Since the triangle is in the 3rd quadrant, all trigonometric ratios, except tan
and cot are negative.
So,cosθ=Base/Hypotenuse Cosθ=-4/5
secθ=Hypotnuse/Base secθ=-5/4
tanθ=Perpendicular/Base tanθ=3/4
cotθ=Base/Perpendicular cotθ=4/3
Answer:
7 large pizzas
Step-by-step explanation:
multiply 9.50 by every number on the calculator and get 9.50×7=66.5
or just divide 75.50 by 9.50
or idk
Answer:
x+5 (under assumption you meant to do -3x
Step-by-step explanation:
you can use long division.
Take the leading coefficient x^4 and divide it by x^3. This results in x which is going to be the first part of you quotient. Now take that x and multiply it by the divisor (x^3 - 3). This gives you x(x^3 - 3) = x^4 - 3x. Now subtract that x^4 - 3x from the original polynomial and repeat this until you can't divide anymore
Answer:
1st one: (x-2)(x+10)
2nd one: (x-7)(x-3)
Step-by-step explanation:
