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rosijanka [135]
3 years ago
5

Remember to show work and explain. Use the math font.

Mathematics
1 answer:
MrMuchimi3 years ago
6 0

Answer:

\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}

Step-by-step explanation:

\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================

1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}

\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)

--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)

--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}

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Step-by-step explanation:

y = 3x   (-3,-9)

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-9 = -9

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Which of the following is equivalent to <br><br> 3x−4y=18?
sammy [17]

The equation which is equivalent to the equation as given in the task content is; 6x -8y = 36.

<h3>Which equation is equivalent to the equation as given in the task content?</h3>

According to the task content, it follows that the equation which is given in the task content is;

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Hence, upon multiplication of the whole equation by 2; the resulting equation from the multiplication is;

6x -8y = 36

Ultimately, the equivalent equation is; 6x -8y = 36.

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8 0
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Please help!
erik [133]
For the answer to the question above, 
1 + nx + [n(n-1)/(2-factorial)](x)^2 + [n(n-1)(n-2)/3-factorial] (x)^3 

<span>1 + nx + [n(n-1)/(2 x 1)](x)^2 + [n(n-1)(n-2)/3 x 2 x 1] (x)^3 </span>

<span>1 + nx + [n(n-1)/2](x)^2 + [n(n-1)(n-2)/6] (x)^3 </span>

<span>1 + 9x + 36x^2 + 84x^3 </span>

<span>In my experience, up to the x^3 is often adequate to approximate a route. </span>

<span>(1+x) = 0.98 </span>

<span>x = 0.98 - 1 = -0.02 </span>

<span>Substituting: </span>

<span>1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 </span>

<span>approximation = 0.834 </span>

<span>Checking the real value in your calculator: </span>

<span>(0.98)^9 = 0.834 </span>

<span>So you have approximated correctly. </span>

<span>If you want to know how accurate your approximation is, write out the result of each in full: </span>

<span>1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 = 0.833728 </span>

<span> (0.98)^9 = 0.8337477621 </span>

<span>So it is correct to 4</span>
5 0
3 years ago
Read 2 more answers
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