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3 years ago

The graph shows two exponential functions. Choose the correct statement concerning their bases

Mathematics

1 answer:

Irina18 [472]3 years ago

4 0

Answer:

Step-by-step explanation:

The two graphs are reflections of each other in the y-axis

If one is f(x), other is f(-x)

c^x = b^(-x)

c = b^-1

c = 1/b

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If f is a differentiable function, find an expression for the derivative of each of the following functions

almond37 [142]

12a. Answer: d) x⁶ f'(x) + 6x⁵ f(x)

<u>Step-by-step explanation:</u>

Use the multiplication formula for derivatives:

y = a · b → y' = a'b + ab'

a = x⁶ b = f(x)

a' = 6x⁵ b' = f'(x)

y' = 6x⁵ f(x) + x⁶ f'(x)

***************************************************************

12b. Answer: $\bold{b)\ y=\dfrac{xf'(x)-9f(x)}{x^{10}}'}$

<u>Step-by-step explanation:</u>

Use the division formula for derivatives:

$y=\dfrac{a}{b}$ → $y' = \dfrac{a'b - ab'}{b^2}$

$y=\dfrac{f(x)}{x^9}\\\\a=f(x)\qquad b=x^9\\\\a'=f(x)\qquad b'=9x^8\\\\y'=\dfrac{a'b-ab'}{b^2}\\\\y'=\dfrac{x^9f'(x)-9x^8f(x)}{(x^9)^2}\\\\.\ =\dfrac{x^9f'(x)-9x^8f(x)}{x^{18}}\\\\\text{factor out }x^{8}: y'=\dfrac{xf'(x)-9f(x)}{x^{10}}$

Note: You can also move the denominator to the top (it will have a negative exponent) and use the multiplication formula for derivatives.

5 0

3 years ago

6th grade math help me pleaseeee

azamat

Answer:

The 1st one

Step-by-step explanation:

3 0

3 years ago

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A growth-mindset researcher plans to take an SRS of 200200200 teenagers from the population of teenagers in North America to see

tresset_1 [31]

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3 years ago

What is the solution of -8/2y-8=5/y+4-7y+8/y^2-16

Mnenie [13.5K]

<h2>Answer:</h2>

The solution of the given equation is:

y=6

<h2>Step-by-step explanation:</h2>

The expression is given by:

$\dfrac{-8}{2y-8}=\dfrac{5}{y+4}-\dfrac{7y+8}{y^2-16}$

Now on solving for the given equation

$\dfrac{-8}{2(y-4)}=\dfrac{5}{y+4}-\dfrac{7y+8}{(y-4)(y+4)}$

since,

$a^2-b^2=(a-b)(a+b)\\\\so,\\\\y^2-16=y^2-4^2\\\\i.e.\\\\y^2-16=(y-4)(y+4)$

Hence, we get:

$\dfrac{-4}{y-4}=\dfrac{5\times (y-4)-(7y+8)}{(y+4)(y-4)}\\\\i.e.\\\\\dfrac{-4}{y-4}=\dfrac{5y-20-7y-8}{(y-4)(y+4)}$

$-4\times (y-4)(y+4)=(-2y-28)(y-4)\\\\i.e.\\\\(y-4)(-4\times (y+4))=(-2y-28)(y-4)\\\\i.e.\\\\(y-4)(-4y+16)=(-2y-28)(y-4)\\\\i.e.$

$(y-4)(-4y+16)-(-2y-28)(y-4)=0\\\\i.e.\\\\(y-4)(-4y-16+2y+28)=0\\\\i.e.\\\\(y-4)(-2y+12)=0\\\\i.e.$

$y-4=0\ or\ -2y+12=0\\\\i.e.\\\\y=4\ or\ 2y=12\\\\i.e.\\\\y=4\ or\ y=6$

but y≠ 4

since, the denominator of the term in the left side of the given equality and the second term in the right side of the given equality will be zero and hence, the expression will be not defined.

Hence, the value of y is: 6

7 0

4 years ago

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What is 6(-3a + 12) + 8a - 13a

Shtirlitz [24]

It is 23a+72
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6(-3a+12)+8a-13a
-18a+72+8a-13a
23a+72
do you understand

5 0

3 years ago

Read 2 more answers