Y= -x and y= -1/x
Are one-one
y= -x, inverse x= -y
Identify function
hence, inverse of y= -x in y= -x
similarly in the case of y= -1/x
so option (D) I and II, only
Step-by-step explanation:
S+T+U=180°{sum of angle of triangle}
- 68°+90°+U=180°
- U=180-158°
- U=22°
sin theta= p/h
sin22°=x/83
o.37=x/83
x=0.37×83
x=30.71
x=31
stay safe healthy and happy.
D.$6,270 you just multiple
Hello:
<span>Use De Moivre’s Theorem :
</span>(3(cos27 +isin27)^5 = 3^5( cos(27 × 5) +isin(27 × 5))
= 3^5 ( cos(135)+i sin(135))
= 3^5(-√2/2+i √2/2)
because : cos(135) = -√2/2 and sin(135) = √2/2
(3(cos27 +isin27)^5 = (- 3^5√2/2)+ i ( 3^5√2/2) ...(form : a+ib when
a= (- 3^5√2/2) and b = ( 3^5√2/2)
(a) By the fundamental theorem of calculus,
<em>v(t)</em> = <em>v(0)</em> + ∫₀ᵗ <em>a(u)</em> d<em>u</em>
The particle starts at rest, so <em>v(0)</em> = 0. Computing the integral gives
<em>v(t)</em> = [2/3 <em>u</em> ³ + 2<em>u</em> ²]₀ᵗ = 2/3 <em>t</em> ³ + 2<em>t</em> ²
(b) Use the FTC again, but this time you want the distance, which means you need to integrate the <u>speed</u> of the particle, i.e. the absolute value of <em>v(t)</em>. Fortunately, for <em>t</em> ≥ 0, we have <em>v(t)</em> ≥ 0 and |<em>v(t)</em> | = <em>v(t)</em>, so speed is governed by the same function. Taking the starting point to be the origin, after 8 seconds the particle travels a distance of
∫₀⁸ <em>v(u)</em> d<em>u</em> = ∫₀⁸ (2/3 <em>u</em> ³ + 2<em>u</em> ²) d<em>u</em> = [1/6 <em>u</em> ⁴ + 2/3 <em>u</em> ³]₀⁸ = 1024
