Can someone plss help me

Mathematics

1 answer:

puteri [66]3 years ago

7 0

0/3 is the same thing as 0, so write those nxt to each other. Then 1/3, 2/3. Lastly 3/3 is the same as 1, so write those nxt to each other.

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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)

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

7 0

3 years ago

Joe would like to paint a rectangular shaped wall. The measures of dimensions of the wall are 8 inches by 4 inches. Every 1 inch

Sergio [31]

Answer:

$90

Step-by-step explanation:

8 + 4 = 12

12 x 3 (3 feet for each inch) = 36

36 x 2.50 (cost per square foot) = $90

3 0

3 years ago

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Find the missing measure for a right circular cone given the following information. Find L.A. if r = 6 and h = 8

Masja [62]

$\bf \textit{lateral area of a cone}\\\\
LA=\pi r\sqrt{r^2+h^2}\quad 
\begin{cases}
r=radius\\
h=height\\
-----\\
r=6\\
h=8
\end{cases}\implies 6\pi \sqrt{6^2+8^2}$