The answer will be 625j^4 k^4
The first one is incorrect. i’d pick that one.
To find it, evaluate it at the endpoints and the vertex
in form
f(x)=ax²+bx+c
the x value of the vertex is -b/2a
given
c(t)=1t²-10t+76
x value of vertex is -(-10)/1=10
evaluate c(0) and c(13) and c(10)
c(0)=76
c(13)=115
c(10)=76
it reached minimum in 2000 and 2010
porbably teacher wants 2010
the min value is $76
The square on the hypotenuse is skew and impossible to visually validate. There's nothing to sum. It is a simple calculation to note that <span><span><span>32</span>+<span>42</span>=<span>52</span></span><span><span>32</span>+<span>42</span>=<span>52</span></span></span><span>, but its picture doesn't 'say' that the angle between the 3 and 4 sided square must be 90 degrees.
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