What is the approximate value of 14 to the nearest tenth?

Mathematics

1 answer:

kykrilka [37]2 years ago

5 0

Answer:

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Explanation:

10 is closer to 14 than 20 and 14 is between 10 and 20, so you can only compare between 10 and 20.

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d1i1m1o1n [39]

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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:

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

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