Can someone please help me with this!!!

Mathematics

1 answer:

user100 [1]3 years ago

7 0

It is an arithmetic because it is adding 2 to each number.

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Write an expression for the calculation the quotient of 16 minus 2 and 7 minus 5

Tpy6a [65]

Answer:

$(16-2)=(7-2)\times7+0\\\\Quotient=7$

Step-by-step explanation:

$we\ have\ to\ find\ quotient\ of\ 16-2\ and\ 7-5\\\\ Quotient=(16-2)\div (7-5)\\\\Quotient=\frac{16-2}{7-5}\\\\Quotient=\frac{14}{2}\\\\Quotient=7\times 2+0\\\\Quotient=7$

8 0

2 years ago

Factor the polynomial 8k^2 +80k +200

blondinia [14]

<h3>Solution:</h3>

$8 {k}^{2} + 80k + 200 \\ = 8( {k}^{2} + 10k + 25) \\ = 8 [{k}^{2} + 2(k)(5) + ({5})^{2} ] \\ = 8( k + 5) ^{2}$

<h3>Answer:</h3>

$8( k + 5) ^{2}$

<h3>Hope it helps..</h3>

ray4918 here to help

6 0

2 years ago

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)

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