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3 years ago

Why is b ≠ 1 a condition for f(x) = abx (x is a exponent i cant put it for some reason)?

Mathematics

1 answer:

timurjin [86]3 years ago

7 0

Answer:

1^x = 1

Step-by-step explanation:

If b = 1,

It will be a horizontal line:

y = a

Not an exponential curve

Because

y = a(1^x)

1 power anything is 1

y = a × 1

y = a

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Solve for x in the triangle,round your answer to the nearest tenth

Nuetrik [128]

Cosine = adjacent / hypotenuse

cos (49)= x/27
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27 * 0.656 = x
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2 years ago

Cos ( α ) = √ 6/ 6 and sin ( β ) = √ 2/4 . Find tan ( α − β )

Zina [86]

Answer:

$\purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$

Step-by-step explanation:

$\cos( \alpha ) = \frac{ \sqrt{6} }{6} = \frac{1}{ \sqrt{6} } \\ \\ \therefore \: \sin( \alpha ) = \sqrt{1 - { \cos}^{2} ( \alpha ) } \\ \\ = \sqrt{1 - \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} } \\ \\ = \sqrt{1 - {\frac{1}{ {6} }}} \\ \\ = \sqrt{ {\frac{6 - 1}{ {6} }}} \\ \\ \red{\sin( \alpha ) = \sqrt{ { \frac{5}{ {6} }}} } \\ \\ \tan( \alpha ) = \frac{\sin( \alpha ) }{\cos( \alpha ) } = \sqrt{5} \\ \\ \sin( \beta ) = \frac{ \sqrt{2} }{4} \\ \\ \implies \: \cos( \beta ) = \sqrt{ \frac{7}{8} } \\ \\ \tan( \beta ) = \frac{\sin( \beta ) }{\cos( \beta ) } = \frac{1}{ \sqrt{7} } \\ \\ \tan( \alpha - \beta ) = \frac{ \tan \alpha - \tan \beta }{1 + \tan \alpha . \tan \beta} \\ \\ = \frac{ \sqrt{5} - \frac{1}{ \sqrt{7} } }{1 + \sqrt{5} . \frac{1}{ \sqrt{7} } } \\ \\ = \frac{ \sqrt{35} - 1 }{ \sqrt{7} + \sqrt{5} } \\ \\ \purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$

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3 years ago

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spin [16.1K]

Answer:

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Step-by-step explanation:

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it can also be written as 29/12

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3 years ago

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choli [55]

<h3>Answer: 5^24</h3>

<h2>Maybe it can be your answer but i am not sure that it is fixed answer </h2>

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