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3 years ago

Simplify this please help the answer choice are in the picture

Mathematics

2 answers:

rjkz [21]3 years ago

6 0

Answer:

6^1/12

Step-by-step explanation:

You can rewrite the numbers as 6^1/3 and 6^1/4. Now we need common denominators on the powers. You can make the 1/3 into 4/12 and the 1/4 into 3/12. When you divide powers, you just subtract the numerators for each fraction. So you do 4-3. This leaves you with 6^1/12. Hope this helps :)

pashok25 [27]3 years ago

3 0

Answer:

6^ (1/12)

Step-by-step explanation:

6 ^ 1/3

-------------

6 ^ 1/4

x^a / x^b = x^ (a-b)

6^(1/3-1/4)

Getting a common denominator)

6^(4/12-3/12)

6^1/12

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Write out the steps that you would follow to solve the equation: 3(x-4) = 2x +5

luda_lava [24]

Answer:

x = 17

Step-by-step explanation:

To solve this equation, we can illustrate the algebraic concepts being employed and solve for the variable.

3(x - 4) = 2x + 5 Distribute the 3 into the parentheses.

3x - 12 = 2x + 5 Subtract 2x from both sides of the equation to combine like terms.

x - 12 = 5 Add 12 to both sides of the equation.

x = 17

6 0

3 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)

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