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2 years ago

Please help me. I'm desperate

Mathematics

1 answer:

Flauer [41]2 years ago

7 0

Ok so x is the original price and we want to solve if the original price was 48 so x=48. When we substitute this into the equation we get 0.75(48)-0.15(0.75x48). When we simplify this expression we get 30.6. The current price is $30.60. Hope this helps.

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2 years ago

Find all exact solutions on [0, 2π). (Enter your answers as a comma-separated list.)

nalin [4]

Answer:

$0, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \pi$

Step-by-step explanation:

Given the equation:

$sin(x)-2sin(x) tan(x)=0$

To find:

The solutions of given equation in the range [0, 2 $\pi$ ) i.e. 0 can be in the answer but $2\pi$ can not be there in the answer.

Solution:

Taking $tan(x)$ common, we get:

$tan(x) (1-2sin(x))=0$

The solutions can be given as:

$tan(x) = 0$ OR $1-2sin(x) = 0$

Let us solve both the equations one by one:

First equation:

$tan(x) = 0\\ \Rightarrow x=0, \pi$

Second equation:

$1-2sin(x) = 0\\\Rightarrow 2sinx=1\\\Rightarrow sinx=\dfrac{1}{2}\\\Rightarrow x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$

Therefore, the answers as a comma separated list are:

$0, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \pi$

7 0

2 years ago

Question regarding logarithms.

Eddi Din [679]

$5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}$

$\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$

$\dfrac{5^2\cdot5^2}{7^3\cdot7}=\left(\dfrac{5}{7}\right)^{x-2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{5^4}{7^4}=\left(\dfrac{5}{7}\right)^{x-2}\\\\\left(\dfrac{5}{7}\right)^4=\left(\dfrac{5}{7}\right)^{x-2}\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}$