Answer:
(C)
Step-by-step explanation:
Given the distance, d(t) of a particle moving in a straight line at any time t is:
To find an expression for the instantaneous velocity v(t) of the particle at any given point in time, we take the derivative of d(t).
The correct option is C.
Let <em>a</em> and <em>b</em> be the zeroes of <em>x</em>² + <em>kx</em> + 12 such that |<em>a</em> - <em>b</em>| = 1.
By the factor theorem, we can write the quadratic in terms of its zeroes as
<em>x</em>² + <em>kx</em> + 12 = (<em>x</em> - <em>a</em>) (<em>x</em> - <em>b</em>)
Expand the right side and equate the coefficients:
<em>x</em>² + <em>kx</em> + 12 = <em>x</em>² - (<em>a</em> + <em>b</em>) <em>x</em> + <em>ab</em>
Then
<em>a</em> + <em>b</em> = -<em>k</em>
<em>ab</em> = 12
The condition that |<em>a</em> - <em>b</em>| = 1 has two cases, so without loss of generality assume <em>a</em> > <em>b</em>, so that |<em>a</em> - <em>b</em>| = <em>a</em> - <em>b</em>.
Then if <em>a</em> - <em>b</em> = 1, we get <em>b</em> = <em>a</em> - 1. Substitute this into the equations above and solve for <em>k</em> :
<em>a</em> + (<em>a</em> - 1) = -<em>k</em> → 2<em>a</em> = 1 - <em>k</em> → <em>a</em> = (1 - <em>k</em>)/2
<em>a</em> (<em>a</em> - 1) = 12 → (1 - <em>k</em>)/2 • ((1 - <em>k</em>)/2 - 1) = 12
→ (1 - <em>k</em>)²/4 - (1 - <em>k</em>)/2 = 12
→ (1 - <em>k</em>)² - 2 (1 - <em>k</em>) = 48
→ (1 - 2<em>k</em> + <em>k</em>²) - 2 (1 - <em>k</em>) = 48
→ <em>k</em>² - 1 = 48
→ <em>k</em>² = 49
→ <em>k</em> = ± √(49) = ±7
An hour has 60 minutes so 1/40 of an hour is:
(1/40)(60) = 60/40 = 6/4 = 3/2 = 1.5
so 1/40 of an hour is a minute and a half
Divide 360 by 24 to find 1%. Then multiply by 100. 360/24 *100= 1500
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c .
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d.
