$\bf \textit{Pythagorean Identities}
\\\\
sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\ 1+tan^2(\theta)=sec^2(\theta)\\\\
-------------------------------\\\\\
[1+tan^2(u)][1-sin^2(u)]=1
\\\\\\\
[sec^2(u)][cos^2(u)]\implies \cfrac{1}{cos^2(u)}\cdot cos^2(u)\implies \cfrac{cos^2(u)}{cos^2(u)}\implies 1$

7 0

3 years ago

Tom and jasper have the same percentage of blue marbles in their bags of marble tom has 5 blue marbles and 20 total marbles jasp

Lesechka [4]

8 of them are not blue

6 0

2 years ago

I'm having trouble with algebraic fractions . 3/2n-8/3=-29/12

nadezda [96]

what we can do is hmmmm get the LCD of all fractions, and multiply both sides by that LCD, let's see, we have denominators of 2, 3 and 12, so we can use the LCD of 12, and multiply both sides by 12 to do away with the denominators.

$\bf \cfrac{3}{2}n-\cfrac{8}{3}=-\cfrac{29}{12}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{12}}{12\left( \cfrac{3}{2}n-\cfrac{8}{3} \right)=12\left( -\cfrac{29}{12} \right)}\implies (6)3n-(4)8=(1)-29 \\\\\\ 18n-32=-29\implies 18n=3\implies n=\cfrac{3}{18}\implies n=\cfrac{1}{6}$

5 0

3 years ago