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Ivenika [448]
3 years ago
7

Solve the exponent equation for x 7(4x-6)= 1/49

Mathematics
1 answer:
tatiyna3 years ago
4 0
X=1. Chose C. Hoped that will help
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The answer is 12


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Y=arccos(1/x)<br><br> Please help me do them all! I don’t know derivatives :(
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Answer:

f(x) =  {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x  \rightarrow \: x = sec \: y\\  \frac{dx}{dx}  =  \frac{d(sec \: y)}{dx}  \\ 1 = \frac{d(sec \: y)}{dx} \times  \frac{dy}{dy}  \\ 1 = \frac{d(sec \: y)}{dy} \times  \frac{dy}{dx}  \\1 = tan \: y.sec \: y. \frac{dy}{dx}  \\ \frac{dy}{dx} =  \frac{1}{tan \: y.sec \: y}  \\ \frac{dy}{dx} =  \frac{1}{ \sqrt{( {sec}^{2}   \: y - 1}) .sec \: y}  \\ \frac{dy}{dx} =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} } \\   \therefore  \frac{d( {sec}^{ - 1}x) }{dx}  =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx}  =  \frac{1}{ |5x |  \sqrt{25 {x }^{2}  - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx}  =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} }

4 0
2 years ago
Write a two column proof given
never [62]
Im sorry but im not sure what your asking
8 0
3 years ago
3/4 divided by 4/5 is 15/16 is it in its lowest terms​
Softa [21]

Answer:

yes it cant be siplified anymore

Step-by-step explanation:

8 0
3 years ago
What is one solution of 5x-13<br>​
Oduvanchick [21]

Answer:

5x-13= 12

Step-by-step explanation:

5x-13 = 12

    +13   +13

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

5x= 25

x=5

5 0
3 years ago
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