$\text{Here,}\\\\\text{Opposite side,}~ O = 2\\\\\text{Hypotenuse,}~ H = \sqrt 5 \\\\\text{Adjacent side,}~ A = \sqrt{5-4} = \sqrt 1 =1 ~~ ;[\text{By using Pythagorean theorem}]\\\\\\\sin \theta = \dfrac{O}H = \dfrac 2{\sqrt 5}\\\\\\\cos \theta = \dfrac{A}{H} = \dfrac{1}{\sqrt 5} \\ \\\\ \tan \theta = \dfrac{O}{A} = \dfrac 21 = 2$

$csc \theta = \dfrac{1}{\sin \theta} = \dfrac{\sqrt 5}2\\\\\\\sec \theta = \dfrac{1}{\cos \theta} = \sqrt 5 \\\\\\\cot \theta = \dfrac 1{\tan \theta} = \dfrac 12$

5 0

3 years ago

Can someone plz help me with Algebra 2....... :) Will give Brainliest!!

Anarel [89]

Using logarithms property of log(x)+log(y)=log(xy)
so here, you can sum the equation to;
$log((x+6)*(x-6))=2$
so you can simply say that;
$log_{8}((x+6)( x-6))=2$
and by multiplying (x+6)*(x-6)
$log_{8}(x^2-36)=2$
and as you know also that;
$a^{b}=c$ is same as $log _{a}c=b$
so you can simply state it as;
$8^2=x^2-36
64=x^2-36
64+36=x^2
x^2=100
x=10$
And you can check your work by substituting with 10 instead of x in the original function.
Hope this helps!

6 0

3 years ago