If the position at time <em>t</em> is
<em>s(t)</em> = (1 m/s³) <em>t</em> ³
then the average velocity over <em>t</em> = 2 s and <em>t</em> = 2.001 s is
<em>v</em> (ave) = (<em>s</em> (2.001 s) - <em>s</em> (2 s)) / (2.001 s - 2 s)
<em>v</em> (ave) = ((1 m/s³) (2.001 s)³ - (1 m/s³) (2 s)³) / (2.001 s - 2 s)
<em>v</em> (ave) ≈ (8.01201 m - 8 m) / (0.001 s)
<em>v</em> (ave) ≈ 12.006 m/s
Answer:
2
Step-by-step explanation:
there would be 2 in every group, 10 divided by 5
The domain is going to be ur x values....the number of minutes
therefore, ur domain is : I know it is either A or C.....I am thinking C because it does not have to be a whole minute...it can be 1.5 minutes...so that would eliminate A. final answer is gonna be C
<span>w^2 + 18w + 77 = 0
breaking the middle term
</span><span>w^2 + 11w + 7w + 77 = 0
making pairs,
</span>(w^2 + 11w) + (7w<span> + 77) = 0
</span>w(w + 11) + 7(w + 11) = 0
(w+11)(w+7)=0
Hence,
the best answer is :
<span>C. (w + 7)(w + 11) </span>
Answer:
x=21° and ∠KMH=96°
Step-by-step explanation:
From the given information that a bike path crosses a road and the given figure, we get
∠GMH=∠KMI=84°(Vertically opposite angles)
⇒4x=84°
⇒x=21°
therefore, the value of ∠GMH=4x=4(21)=84°
Now, ∠KMH+∠KMI=180°( Linear pair)
⇒∠KMH+84°=180°
⇒∠KMH=180-84
⇒∠KMH=96°
Thus, the value of ∠KMH is 96°
