23÷5=4 with a remainder of $3. each avocada was 4$.
Step 1: Line up the equations so that the variables are lined up vertically.
Step 2: Choose the easiest variable to eliminate and multiply both equations by different numbers so that the coefficients of that variable are the same.
Step 3: Subtract the two equations.
Step 4: Solve the one variable system.
Step 5: Put that value back into either equation to find the other equation.
Step 6: Reread the question and plug your answers back in to check.
Answer:
see explanation
Step-by-step explanation:
The amplitude is the value from the x- axis to the maximum, thus
amplitude = 2
the midline is the line midway between the minimum and the maximum
Assuming the graph is symmetrical about the x- axis
Then the max is 2 and the min is - 2
y = 
= 
= 0
Thus the midline is y = 0
keeping in mind that perpendicular lines have <u>negative reciprocal</u> slopes, hmmm what's the slope of y=2/3x-1 anyway?
![\bf \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}~\hspace{7em}y=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{2}{3}}x-1 \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D~%5Chspace%7B7em%7Dy%3D%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B%5Ccfrac%7B2%7D%7B3%7D%7Dx-1%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
so, notice, "one of his mistakes" is that he used 3/2 as the slope, not -3/2.
so, we're really looking for a line whose slope is -3/2 and runs through (-7, 1/2).
![\bf (\stackrel{x_1}{-7}~,~\stackrel{y_1}{\frac{1}{2}})~\hspace{10em} slope = m\implies -\cfrac{3}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\cfrac{1}{2}=-\cfrac{3}{2}[x-(-7)]\implies y-\cfrac{1}{2}=-\cfrac{3}{2}(x+7) \\\\\\ y-\cfrac{1}{2}=-\cfrac{3}{2}x-\cfrac{21}{2}\implies y=-\cfrac{3}{2}x-\cfrac{21}{2}+\cfrac{1}{2}\implies y=-\cfrac{3}{2}x-10](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B-7%7D~%2C~%5Cstackrel%7By_1%7D%7B%5Cfrac%7B1%7D%7B2%7D%7D%29~%5Chspace%7B10em%7D%20slope%20%3D%20m%5Cimplies%20-%5Ccfrac%7B3%7D%7B2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-%5Ccfrac%7B1%7D%7B2%7D%3D-%5Ccfrac%7B3%7D%7B2%7D%5Bx-%28-7%29%5D%5Cimplies%20y-%5Ccfrac%7B1%7D%7B2%7D%3D-%5Ccfrac%7B3%7D%7B2%7D%28x%2B7%29%20%5C%5C%5C%5C%5C%5C%20y-%5Ccfrac%7B1%7D%7B2%7D%3D-%5Ccfrac%7B3%7D%7B2%7Dx-%5Ccfrac%7B21%7D%7B2%7D%5Cimplies%20y%3D-%5Ccfrac%7B3%7D%7B2%7Dx-%5Ccfrac%7B21%7D%7B2%7D%2B%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20y%3D-%5Ccfrac%7B3%7D%7B2%7Dx-10)
Input is 1,2,3,4,5,6 and output is 5,8,11,14,17,20 what is the rule and equation
Alecsey [184]
Notice that each output is increased by 3 and starts with 5. We can predict that when input is 0, the output would be 2.
So the rule would be multiply the input by 3 then add 2. The equation would be f(n) = 3n+2
