Solve -3x + 12 < 18 help-

Mathematics

2 answers:

Colt1911 [192]2 years ago

7 0

Answer:

x>−2

Step-by-step explanation:

−3x+12−12<18−12

−3x<6

Step 2: Divide both sides by -3.

−3x

−3

Mademuasel [1]2 years ago

6 0

X>-2 I’m pretty sure this is right. Have a great day!

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Find T, N, and kappa for the plane curve Bold r left parenthesis t right parenthesis equalsleft parenthesis 7 Bold cos t plus

pickupchik [31]

Answer:

Step-by-step explanation:

r(t) = (7 cost + 7t sin t)i + (7 sin t - 7t cos t)j

$\frac{d \bar r t}{dt} =(7\frac{d}{dt}\cos t + 7\frac{d}{dt} (t \sin t)i+(7\frac{d}{dt} \sin t-7\frac{d}{dt} t \cos t)j$

$=(7(-\sin t)+7(1* \sin t+t \cos t))i+(7 \cost -7(1*\cos t - t \sin t))j\\\\=7((-\sin t+\sin t+t \cos t)i+(\cos t-\cos t+t \sin t)j)\\\\=7((t\cos t)i+(t\sin t)j)$

$\bar r'(t)=\frac{d \bar r t}{dt} =(7t\cos t)i+(7t\sin t)j---(1)\\\\11\bar r(t)=\sqrt{(7t\cos t)^2+(7t\sin t)^2}\\\\=\sqrt{49t^2(\cos^2t+\sin^2 t)} \\\\=7t$

$\bar T (t)=\frac{\bar r'(t)}{11\bar r(t)11} =\frac{(7t\cos t)i+(7t\sin t)j}{7t} \\\\\barT(t)=(\cos t)i+(\sin t)j$

$\bar T'(t)=\frac{d}{dt} (\cos t)i+\frac{d}{dt} (\sin t) j\\\\\bar T'(t)=(-\sin t)i+(\cos t)j---(2)\\\\11\bar T'(t)=\sqrt{(-\sin t)^2+(\cos t)^2} \\\\=\sqrt{\sin^2t+\cos^2t} \\\\=1$

$\bar N(t)=\bar T'(t)=\frac{(-\sin t)i+(\cos t)j}{(1)} \\\\ \large \boxed {\bar N(t)=(-\sin t)i+(\cos t)j}$

$K(t)=\frac{|\b\r T'(t)|}{\bar r (t)|} \\\\=\frac{|-\sin t i+\cos t j|}{|7t\cos t +7t \sin t j|}$

Using eq (1) and (2)

$K(t)=\frac{\sqrt{(-\sin t)^2+(\cos t)^2} }{\sqrt{(7t\cos t)^2+(7t\sin t)^2} }\\\\=\frac{\sqrt{\sin^2 t+\cos^2t} }{\sqrt{49t^2(\cos^2 t+\sin^2t)} }\\\\=\frac{\sqrt{1} }{\sqrt{49t^2\times 1} } \\\\ \large \boxed {K(t)=\frac{1}{7t} }$