Answer: ![\bold{5)\ \cos \theta=\dfrac{\sqrt{11}}{6}\qquad 6)\ \tan \theta =\dfrac{8}{17}\qquad 7)\ \cos \theta = \dfrac{4}{3}\qquad 8)\ \cos \theta = \dfrac{\sqrt{10}}{10}}](https://tex.z-dn.net/?f=%5Cbold%7B5%29%5C%20%5Ccos%20%5Ctheta%3D%5Cdfrac%7B%5Csqrt%7B11%7D%7D%7B6%7D%5Cqquad%206%29%5C%20%5Ctan%20%5Ctheta%20%3D%5Cdfrac%7B8%7D%7B17%7D%5Cqquad%207%29%5C%20%5Ccos%20%5Ctheta%20%3D%20%5Cdfrac%7B4%7D%7B3%7D%5Cqquad%208%29%5C%20%5Ccos%20%5Ctheta%20%3D%20%5Cdfrac%7B%5Csqrt%7B10%7D%7D%7B10%7D%7D)
<u>Step-by-Step Explanation:</u>
Pythagorean Theorem is: a² + b² = c² , <em>where "c" is the hypotenuse</em>
![5)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{3\sqrt{11}}{18}\quad \rightarrow \large\boxed{\dfrac{\sqrt{11}}{6}}](https://tex.z-dn.net/?f=5%29%5C%20%5Ccos%20%5Ctheta%3D%5Cdfrac%7B%5Ctext%7Bside%20adjacent%20to%7D%5C%20%5Ctheta%7D%7B%5Ctext%7Bhypotenuse%20of%20triangle%7D%7D%3D%5Cdfrac%7B3%5Csqrt%7B11%7D%7D%7B18%7D%5Cquad%20%5Crightarrow%20%5Clarge%5Cboxed%7B%5Cdfrac%7B%5Csqrt%7B11%7D%7D%7B6%7D%7D)
Note: (15)² + (3√11)² = hypotenuse² → hypotenuse = 18
![6)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{8}{17}\quad =\large\boxed{\dfrac{8}{17}}](https://tex.z-dn.net/?f=6%29%5C%20%5Ccos%20%5Ctheta%3D%5Cdfrac%7B%5Ctext%7Bside%20adjacent%20to%7D%5C%20%5Ctheta%7D%7B%5Ctext%7Bhypotenuse%20of%20triangle%7D%7D%3D%5Cdfrac%7B8%7D%7B17%7D%5Cquad%20%3D%5Clarge%5Cboxed%7B%5Cdfrac%7B8%7D%7B17%7D%7D)
Note: 8² + 15² = hypotenuse² → hypotenuse = 17
![7)\ \tan \theta=\dfrac{\text{side opposite to}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{20}{15}\quad \rightarrow \large\boxed{\dfrac{4}{3}}](https://tex.z-dn.net/?f=7%29%5C%20%5Ctan%20%5Ctheta%3D%5Cdfrac%7B%5Ctext%7Bside%20opposite%20to%7D%5C%20%5Ctheta%7D%7B%5Ctext%7Bside%20adjacent%20to%7D%5C%20%5Ctheta%7D%3D%5Cdfrac%7B20%7D%7B15%7D%5Cquad%20%5Crightarrow%20%5Clarge%5Cboxed%7B%5Cdfrac%7B4%7D%7B3%7D%7D)
Note: hypotenuse not needed for tan
![8)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{2}{2\sqrt{10}}\quad =\large\boxed{\dfrac{\sqrt{10}}{10}}](https://tex.z-dn.net/?f=8%29%5C%20%5Ccos%20%5Ctheta%3D%5Cdfrac%7B%5Ctext%7Bside%20adjacent%20to%7D%5C%20%5Ctheta%7D%7B%5Ctext%7Bhypotenuse%20of%20triangle%7D%7D%3D%5Cdfrac%7B2%7D%7B2%5Csqrt%7B10%7D%7D%5Cquad%20%3D%5Clarge%5Cboxed%7B%5Cdfrac%7B%5Csqrt%7B10%7D%7D%7B10%7D%7D)
Note: 2² + 6² = hypotenuse² → hypotenuse = 2√10
3.5 times as much flour as chocolate chips....
f = 3.5c
f = 1 3/4 (or 1.75)
1.75 = 3.5c
1.75 / 3.5 = c
0.5 = c......chocolate chips needed would be 1/2 a cup
Given:
Two numbers
and 2.
To find:
How many
are in 2.
Solution:
Let x be number of
in 2.
![\dfrac{3}{4}\times x=2](https://tex.z-dn.net/?f=%5Cdfrac%7B3%7D%7B4%7D%5Ctimes%20x%3D2)
Divide both sides by
.
![x=\dfrac{2}{\dfrac{3}{4}}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B2%7D%7B%5Cdfrac%7B3%7D%7B4%7D%7D)
![x=2\times \dfrac{4}{3}](https://tex.z-dn.net/?f=x%3D2%5Ctimes%20%5Cdfrac%7B4%7D%7B3%7D)
![x=\dfrac{8}{3}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B8%7D%7B3%7D)
Therefore, required multiplication equation is
and the required division equation is
.
Answer:
66 galletas
Step-by-step explanation:
Si 3/11 de una caja contiene 18 galletas entonces matematicamente significa que:
![\frac{3}{11}g=18](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B11%7Dg%3D18)
Donde g son las galletas, ahora vamos a aislar la variable g:
![\frac{3}{11}g=18\\\\(\frac{11}{3}) \frac{3}{11} g=18 (\frac{11}{3})\\\\g= \frac{198}{3}=66](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B11%7Dg%3D18%5C%5C%5C%5C%28%5Cfrac%7B11%7D%7B3%7D%29%20%5Cfrac%7B3%7D%7B11%7D%20g%3D18%20%28%5Cfrac%7B11%7D%7B3%7D%29%5C%5C%5C%5Cg%3D%20%5Cfrac%7B198%7D%7B3%7D%3D66)
Entonces una caja contiene 66 galletas
E or any letter in the alphabet is known as a variable which is what you need to solve.