The value of f[ -4 ] and g°f[-2] are
and 13 respectively.
<h3>What is the value of f[-4] and g°f[-2]?</h3>
Given the function;
![f(x) = \frac{3x-2}{x+1}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B3x-2%7D%7Bx%2B1%7D)
![g(x)=x+5](https://tex.z-dn.net/?f=g%28x%29%3Dx%2B5)
- f[ -4 ] = ?
- g°f[ -2 ] = ?
For f[ -4 ], we substitute -4 for every variable x in the function.
![f(x) = \frac{3x-2}{x+1}\\\\f(-4) = \frac{3(-4)-2}{(-4)+1}\\\\f(-4) = \frac{-12-2}{-4+1}\\\\f(-4) = \frac{-14}{-3}\\\\f(-4) = \frac{14}{3}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%5C%5C%5C%5Cf%28-4%29%20%3D%20%5Cfrac%7B3%28-4%29-2%7D%7B%28-4%29%2B1%7D%5C%5C%5C%5Cf%28-4%29%20%3D%20%5Cfrac%7B-12-2%7D%7B-4%2B1%7D%5C%5C%5C%5Cf%28-4%29%20%3D%20%5Cfrac%7B-14%7D%7B-3%7D%5C%5C%5C%5Cf%28-4%29%20%3D%20%5Cfrac%7B14%7D%7B3%7D)
For g°f[-2]
g°f[-2] is expressed as g(f(-2))
![g(\frac{3x-2}{x+1}) = (\frac{3x-2}{x+1}) + 5\\\\g(\frac{3x-2}{x+1}) = \frac{3x-2}{x+1} + \frac{5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) = \frac{3x-2+5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) = \frac{8x+3}{x+1}\\\\We\ substitute \ in \ [-2] \\\\g(\frac{3x-2}{x+1}) = \frac{8(-2)+3}{(-2)+1}\\\\g(\frac{3x-2}{x+1}) = \frac{-16+3}{-2+1}\\\\g(\frac{3x-2}{x+1}) = \frac{-13}{-1}\\\\g(\frac{3x-2}{x+1}) = 13](https://tex.z-dn.net/?f=g%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%20%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%2B%205%5C%5C%5C%5Cg%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%20%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%20%2B%20%5Cfrac%7B5%28x%2B1%29%7D%7Bx%2B1%7D%5C%5C%5C%5Cg%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%20%5Cfrac%7B3x-2%2B5%28x%2B1%29%7D%7Bx%2B1%7D%5C%5C%5C%5Cg%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%20%5Cfrac%7B8x%2B3%7D%7Bx%2B1%7D%5C%5C%5C%5CWe%5C%20substitute%20%5C%20in%20%5C%20%5B-2%5D%20%5C%5C%5C%5Cg%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%20%5Cfrac%7B8%28-2%29%2B3%7D%7B%28-2%29%2B1%7D%5C%5C%5C%5Cg%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%20%5Cfrac%7B-16%2B3%7D%7B-2%2B1%7D%5C%5C%5C%5Cg%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%20%5Cfrac%7B-13%7D%7B-1%7D%5C%5C%5C%5Cg%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%2013)
Therefore, the value of f[ -4 ] and g°f[-2] are
and 13 respectively.
Learn more about composite functions here: brainly.com/question/20379727
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(X+5)^2 = x^2 +10x + 25
the expression that is equivalent to
To multiply it we apply FOIL method
Multiply x with x+5 first and then we multiply 5 with x+5
So Katelyn bought 5.68 pounds of peaches. And one pound of peaches cost $1.49. So that would mean that we would have to multiply the pounds of peaches she bought, with how much each pound costs. 5.68 × 1.49 = ?
Once you multiply these two decimals, you would get 8.4632. So she spent $8.4632 on 5.68 pounds of peaches. You would multiply the two decimals to find how much she spent in all.
![\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ 3 &,& -4~) % (c,d) &&(~ 5 &,& 1~) \end{array} \\\\\\ % slope = m slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-(-4)}{5-3}\implies \cfrac{1+4}{5-3}\implies \cfrac{5}{2} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-4)=\cfrac{5}{2}(x-3)\implies y+4=\cfrac{5}{2}x-\cfrac{15}{2}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%203%20%26%2C%26%20-4~%29%20%0A%25%20%20%28c%2Cd%29%0A%26%26%28~%205%20%26%2C%26%201~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20slope%20%20%3D%20m%0Aslope%20%3D%20%20m%5Cimplies%20%0A%5Ccfrac%7B%5Cstackrel%7Brise%7D%7B%20y_2-%20y_1%7D%7D%7B%5Cstackrel%7Brun%7D%7B%20x_2-%20x_1%7D%7D%5Cimplies%20%5Ccfrac%7B1-%28-4%29%7D%7B5-3%7D%5Cimplies%20%5Ccfrac%7B1%2B4%7D%7B5-3%7D%5Cimplies%20%5Ccfrac%7B5%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20point-slope%20intercept%0A%5Cstackrel%7B%5Ctextit%7Bpoint-slope%20form%7D%7D%7By-%20y_1%3D%20m%28x-%20x_1%29%7D%5Cimplies%20y-%28-4%29%3D%5Ccfrac%7B5%7D%7B2%7D%28x-3%29%5Cimplies%20y%2B4%3D%5Ccfrac%7B5%7D%7B2%7Dx-%5Ccfrac%7B15%7D%7B2%7D)
![\bf y=\cfrac{5}{2}x-\cfrac{15}{2}-4\implies y=\cfrac{5}{2}x-\cfrac{23}{2}\impliedby \begin{array}{llll} \textit{now let's multiply both}\\ \textit{sides by }\stackrel{LCD}{2} \end{array} \\\\\\ 2(y)=2\left( \cfrac{5}{2}x-\cfrac{23}{2} \right)\implies 2y=5x-23\implies \stackrel{standard~form}{-5x+2y=-23} \\\\\\ \textit{and if we multiply both sides by -1}\qquad 5x-2y=23](https://tex.z-dn.net/?f=%5Cbf%20y%3D%5Ccfrac%7B5%7D%7B2%7Dx-%5Ccfrac%7B15%7D%7B2%7D-4%5Cimplies%20y%3D%5Ccfrac%7B5%7D%7B2%7Dx-%5Ccfrac%7B23%7D%7B2%7D%5Cimpliedby%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Bnow%20let%27s%20multiply%20both%7D%5C%5C%0A%5Ctextit%7Bsides%20by%20%7D%5Cstackrel%7BLCD%7D%7B2%7D%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A2%28y%29%3D2%5Cleft%28%20%5Ccfrac%7B5%7D%7B2%7Dx-%5Ccfrac%7B23%7D%7B2%7D%20%5Cright%29%5Cimplies%202y%3D5x-23%5Cimplies%20%5Cstackrel%7Bstandard~form%7D%7B-5x%2B2y%3D-23%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Band%20if%20we%20multiply%20both%20sides%20by%20-1%7D%5Cqquad%205x-2y%3D23)
side note: multiplying by the LCD of both sides is just to get rid of the denominators
Answer:
Step-by-step explanation:
<u><em>Not congruent</em></u>
Better to say that there is not enough information to say, "Triangles are congruent"