Answer:
12 units
Step-by-step explanation:
To find any side length of a right triangle given two side lengths, you would use the pythagorean theorem which is a² + b² = c². In this equation, a and b represents either the height of base of the triangle, and c represents the hypotenuse of the triangle (the diagonal line - in your question it is 13). By plugging in the base and hypotenuse, you get the equation of 5² + b² = 13². 5² is equal to 25 and 13² is equal to 169, so the equation is 25 + b² = 169. Subtract 25 for both sides of the equation and you find that b² = 144. Square both sides and the missing length is 12 because √144 = 12.
Some fractions equivalent to 8/3 are 16/6, 24/9, and 32/12.
Answer:
<em>The answer is 0</em>
<em>So I think no solutions</em>
Step-by-step explanation:
You gotta get each equation into slope-intercept form, nd when u do that 9x - 3y = 6 turns into y = 3x - 2, nd 5y = 15x + 10 turns into y = 3x + 2. Add the equations together, nd get 0.
Answer:
Prime factorization: 825 = 3 × 5 × 5 × 11, which can be written 825 = 3 × 5² × 11.
Step-by-step explanation:
Hope this helps!
Step-by-step explanation:
According to this trigonometric function, −C gives you the OPPOSITE terms of what they really are, so be EXTREMELY CAREFUL:
![\displaystyle Phase\:[Horisontal]\:Shift → \frac{-\frac{2}{3}π}{2} = -\frac{π}{3} \\ Period → \frac{2}{2}π = π](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Phase%5C%3A%5BHorisontal%5D%5C%3AShift%20%E2%86%92%20%5Cfrac%7B-%5Cfrac%7B2%7D%7B3%7D%CF%80%7D%7B2%7D%20%3D%20-%5Cfrac%7B%CF%80%7D%7B3%7D%20%5C%5C%20Period%20%E2%86%92%20%5Cfrac%7B2%7D%7B2%7D%CF%80%20%3D%20%CF%80)
Therefore we have our answer.
Extended Information on the trigonometric function
![\displaystyle Vertical\:Shift → D \\ Phase\:[Horisontal]\:Shift → \frac{C}{B} \\ Period → \frac{2}{B}π \\ Amplitude → |A|](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Vertical%5C%3AShift%20%E2%86%92%20D%20%5C%5C%20Phase%5C%3A%5BHorisontal%5D%5C%3AShift%20%E2%86%92%20%5Cfrac%7BC%7D%7BB%7D%20%5C%5C%20Period%20%E2%86%92%20%5Cfrac%7B2%7D%7BB%7D%CF%80%20%5C%5C%20Amplitude%20%E2%86%92%20%7CA%7C)
NOTE: Sometimes, your vertical shift might tell you to shift your graph below or above the <em>midline</em><em> </em>where the amplitude is.
I am joyous to assist you anytime.
