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Tom [10]
3 years ago
11

Find the function rule

Mathematics
2 answers:
Olenka [21]3 years ago
7 0
-5/1
Hope this helps...
belka [17]3 years ago
7 0
5/1
Hope I can help and have great day;)
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2logx=3-2log(x+3) solve for x​
AVprozaik [17]

Answer:

\large\boxed{x=\dfrac{-3+\sqrt{40+10\sqrt{10}}}{2}}

Step-by-step explanation:

2\log x=3-2\log(x+3)\\\\Domain:\ x>0\ \wedge\ x+3>0\to x>-3\\\\D:x>0\\============================\\2\log x=3-2\log(x+3)\qquad\text{add}\ 2\log(x+3)\ \text{to both sides}\\\\2\log x+2\log(x+3)=3\qquad\text{divide both sides by 2}\\\\\log x+\log(x+3)=\dfrac{3}{2}\qquad\text{use}\ \log_ab+\log_ac=\log_a(bc)\\\\\log\bigg(x(x+3)\bigg)=\dfrac{3}{2}\qquad\text{use the de}\text{finition of a logarithm}\\\\x(x+3)=10^\frac{3}{2}\qquad\text{use the distributive property}

x^2+3x=10^{1\frac{1}{2}}\\\\x^2+3x=10^{1+\frac{1}{2}}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\x^2+3x=10\cdot10^\frac{1}{2}\qquad\text{use}\ \sqrt[n]{a}=a^\frac{1}{n}\\\\x^2+3x=10\sqrt{10}\qquad\text{subtract}\ 10\sqrt{10}\ \text{from both sides}\\\\x^2+3x-10\sqrt{10}=0\\\\\text{Use the quadratic formula}\\\\ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\a=1,\ b=3,\ c=-10\sqrt{10}\\\\b^2-4ac=3^2-4(1)(-10\sqrt{10})=9+40\sqrt{10}\\\\x=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2(1)}=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2}\\\\x=\dfrac{-3-\sqrt{10+10\sqrt{10}}}{2}\notin D

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3 years ago
PLEASE HELP ME ASAP. Thanks
Monica [59]
67 Animals. Hope this helps!
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Please help me!! i have no clue what i am doing on this one.
Sonja [21]

Answer:

yr answer is G. 21

<h2>stay safe healthy and happy<u>.</u><u>.</u></h2>

8 0
3 years ago
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What is the inverse of the function f(x) = 2x - 10?
ollegr [7]

Answer:

h(x) = x/2 + 5

Step-by-step explanation:

f(x) = 2x - 10

Step 1. Replace f(x) with y.

y = 2x - 10

Step 2. Switch variables x and y.

x = 2y - 10

Step 3. Solve for y.

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2y = x + 10

y = x/2 + 5

Step 4. Replace y with h(x)

h(x) = x/2 + 5

4 0
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