Step-by-step explanation:
The parent function of the functions of the form f(x)=√x−a+b is f(x)=√x . Note that the domain of f(x)=√x is x≥0 and the range is y≥0 . The graph of f(x)=√x−a+b can be obtained by translating the graph of f(x)=√x to a units to the right and then b units up.
I hope it's helpful!
Answer:
The final answers are x = 1 OR x = -3.
Step-by-step explanation:
Given the equation is x^2 -3 = -2x
Rewriting it in quadratic form as:- x^2 +2x -3 = 0.
a = 1, b = 2, c = -3.
Using Quadratic formula as follows:- x = ( -b ± √(b² -4ac) ) / (2a)
x = ( -2 ± √(4 -4*1*-3) ) / (2*1)
x = ( -2 ± √(4 +12) ) / (2)
x = ( -2 ± √(16) ) / (2)
x = ( -2 ± 4 ) / (2)
x = (-2+4) / (2) OR x = (-2-4) / (2)
x = 2/2 OR x = -6/2
x = 1 OR x = -3
Hence, final answers are x = 1 OR x = -3.
Answer:
domain = 0 < x < infinity / range = y < 4
Step-by-step explanation:
In this equation, you have to treat the number in the bracket first on the basis of BODMAS
15 - [-3]- 4
Note that when two minuses come together the product is a plus sign.
15 +3 - 4
You have to add before you subract
18 - 4 =14
Therefore, 15- [-3] - 4 = 14.
well, keeping in mind that a year has 12 months, that means that 8 months is 8/12 of a year, when Mrs Rojas pull her money out.
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ t=years\to \frac{8}{12}\dotfill &\frac{2}{3} \end{cases} \\\\\\ A=6000[1+(0.04)(\frac{2}{3})]\implies A=6000\left( \frac{77}{75} \right)\implies A=6160](https://tex.z-dn.net/?f=~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%20%5C%246000%5C%5C%20r%3Drate%5Cto%204%5C%25%5Cto%20%5Cfrac%7B4%7D%7B100%7D%5Cdotfill%20%260.04%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B8%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B2%7D%7B3%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D6000%5B1%2B%280.04%29%28%5Cfrac%7B2%7D%7B3%7D%29%5D%5Cimplies%20A%3D6000%5Cleft%28%20%5Cfrac%7B77%7D%7B75%7D%20%5Cright%29%5Cimplies%20A%3D6160)
well, she put in 6000 bucks, got back 160 extra, that's the interest earned in the 8 months.
what if she had left her money for 1 whole year, then
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ t=years\dotfill &1 \end{cases} \\\\\\ A=6000[1+(0.04)(1)]\implies A=6240](https://tex.z-dn.net/?f=~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%20%5C%246000%5C%5C%20r%3Drate%5Cto%204%5C%25%5Cto%20%5Cfrac%7B4%7D%7B100%7D%5Cdotfill%20%260.04%5C%5C%20t%3Dyears%5Cdotfill%20%261%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D6000%5B1%2B%280.04%29%281%29%5D%5Cimplies%20A%3D6240)
so had she left it in for a year, she'd have gotten 6240, namely 240 in interest, well, what fraction of a year's interest was earned? or worded differently, what fraction is 160(8 months) of 240(1 year)?
