Find the zeros<span> of Function (</span>x)= (x+3)^2(x-5)^6 and state the multiplicity. ... (x--5)^6<span> = 0=> </span>x--5<span>= 0 => </span>x<span> = </span>5<span>. Therefore real </span>zero's of the function<span> are --</span>3<span>,5.</span><span />
Answer:
- time: t = -0.3
- minimum: v = 0.55
Step-by-step explanation:
For quadratic ax^2 + bx + c, the extreme value is found at x=-b/(2a). For your quadratic, the minimum is found at ...
t = -(3)/(2(5))
t = -0.3 . . . . . time of minimum velocity
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The value of velocity at that time is ...
v = 5(-0.3)^2 +3(-0.3) +1 = 5(.09) -.9 +1
v = 0.55 . . . . . value of minimum velocity
Answer:
-2.6, -0.4, 0, 0.8, 1.2
Step-by-step explanation:
ascending just means smallest to biggest, so start with the lowest negative and work up
Answer:
A
Step-by-step explanation:
i think