The formula d=rt relates three quantities distance (d) rate (r) and time (t)
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Finding the upper and lower bounds for a definite integral without an equation is pretty hard because how can we find the upper and lower bounds of definite integral if there is no equation given. But I will teach you how to find the lower and upper bounds of a definite integral, when the equation is like this
So, i integrate this,
I know I have a minimum at x=3 because;
f(t )= t^2 − 6t + 11
f′(t) = 2
t−6 = 0
2(t−3) = 0
t = 3
f(5) = 4
f(1) = −4
So, i integrate this,
I know I have a minimum at x=3 because;
f(t )= t^2 − 6t + 11
f′(t) = 2
t−6 = 0
2(t−3) = 0
t = 3
f(5) = 4
f(1) = −4
Answer:
Step-by-step explanation:
Perpendicular equations have OPPOCITE MULTIPLICATIVE INVERCE <em>RATE OF</em><em> </em><em>CHANGES</em><em> </em>[<em>SLOPES</em>], so 1⅕ becomes −⅚, and we move forward with plugging the information into the Slope-Intercept formula:
To write this in Linear Standard Form, perfourm he following:
y = −⅚x + 12
+ ⅚x + ⅚x
___________
⅚x + y = 12 [We cannot leave the equation this way, so multiply the equation by the denominatour to eradicate the fraction.]
6[⅚x + y = 12]
I am joyous to assist you at any time.