Good morning ☕️
Answer:
<h2>x ≥ 2</h2>
Step-by-step explanation:
our expression is defined when x ≥ 1 ,
because , when x ≥ 1
x - 1 ≥ 0
x + 2√(x - 1) ≥ 0
x - 2√(x - 1) ≥ 0
![Calculations:\\\\(\sqrt{x+2\sqrt{x-1} } -\sqrt{x-2\sqrt{x-1} } )^{2} =2^2\\\\then\\\\(x+2\sqrt{x-1} ) +(x-2\sqrt{x-1} ) -2(\sqrt{x+2\sqrt{x-1} })(\sqrt{x-2\sqrt{x-1} } )=4\\\\then\\\\2x - 2\sqrt{(x+2\sqrt{x-1} )(x-2\sqrt{x-1} ) }=4 \\\\then\\\\2x -2\sqrt{(x-2)^2} = 4\\\\then\\\\2x -2|x - 2| =4\\\\then\\\\x - |x-2|=2\\\\now,\\\\case1 : (x\geq2)\\ x - |x-2|=2 \\means \\x-(x-2)=2\\means\\2=2\\\\case2: (1 \leq x \leq2)\\x - |x-2|=2 \\means \\x-(2-x)=2\\means\\2x-2=2\\means\\x=2](https://tex.z-dn.net/?f=Calculations%3A%5C%5C%5C%5C%28%5Csqrt%7Bx%2B2%5Csqrt%7Bx-1%7D%20%7D%20-%5Csqrt%7Bx-2%5Csqrt%7Bx-1%7D%20%7D%20%29%5E%7B2%7D%20%3D2%5E2%5C%5C%5C%5Cthen%5C%5C%5C%5C%28x%2B2%5Csqrt%7Bx-1%7D%20%29%20%2B%28x-2%5Csqrt%7Bx-1%7D%20%29%20-2%28%5Csqrt%7Bx%2B2%5Csqrt%7Bx-1%7D%20%7D%29%28%5Csqrt%7Bx-2%5Csqrt%7Bx-1%7D%20%7D%20%29%3D4%5C%5C%5C%5Cthen%5C%5C%5C%5C2x%20-%202%5Csqrt%7B%28x%2B2%5Csqrt%7Bx-1%7D%20%29%28x-2%5Csqrt%7Bx-1%7D%20%29%20%7D%3D4%20%5C%5C%5C%5Cthen%5C%5C%5C%5C2x%20-2%5Csqrt%7B%28x-2%29%5E2%7D%20%3D%204%5C%5C%5C%5Cthen%5C%5C%5C%5C2x%20-2%7Cx%20-%202%7C%20%3D4%5C%5C%5C%5Cthen%5C%5C%5C%5Cx%20-%20%7Cx-2%7C%3D2%5C%5C%5C%5Cnow%2C%5C%5C%5C%5Ccase1%20%3A%20%28x%5Cgeq2%29%5C%5C%20x%20-%20%7Cx-2%7C%3D2%20%5C%5Cmeans%20%5C%5Cx-%28x-2%29%3D2%5C%5Cmeans%5C%5C2%3D2%5C%5C%5C%5Ccase2%3A%20%281%20%5Cleq%20x%20%5Cleq2%29%5C%5Cx%20-%20%7Cx-2%7C%3D2%20%5C%5Cmeans%20%5C%5Cx-%282-x%29%3D2%5C%5Cmeans%5C%5C2x-2%3D2%5C%5Cmeans%5C%5Cx%3D2)
According to case 1 and 2 the solution set for this equality is {x real number where x ≥ 2 }.
4 inches of rain would fall in 6 hours
<h3><u>Solution:</u></h3>
Given that, A rain storm came through Clifton park
And it was accumulating
inches of rain/hour
So amount of rain accumulated in 1 hour = ![\frac{2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B3%7D)
Thus amount of rain accumulated in six hours is calculated by multiplying the amount of water accumulating per hour and 6
Amount of water accumulated in 6 hours = Amount of water accumulated in 1 hour
6
![\text { Amount of water accumaulated in 6 hours }=\frac{2}{3} \times 6=4](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Amount%20of%20water%20accumaulated%20in%206%20hours%20%7D%3D%5Cfrac%7B2%7D%7B3%7D%20%5Ctimes%206%3D4)
<em><u>Another way:</u></em>
Let "n" be the amount of rain accumulated in 6 hours
1 hour ⇒
rain accumulated
6 hours ⇒ "n"
By cross multiplication, we get
![6 \times \frac{2}{3} = 1 \times n\\\\n = \frac{2}{3} \times 6 = 4](https://tex.z-dn.net/?f=6%20%5Ctimes%20%5Cfrac%7B2%7D%7B3%7D%20%3D%201%20%5Ctimes%20n%5C%5C%5C%5Cn%20%3D%20%5Cfrac%7B2%7D%7B3%7D%20%5Ctimes%206%20%3D%204)
Hence, 4 inches of rain would fall in 6 hours.
Answer:
q = -13
Step-by-step explanation:
The given equation is:
![2x^{2}-8x=5](https://tex.z-dn.net/?f=2x%5E%7B2%7D-8x%3D5)
Taking 2 as common from left hand side, we get:
![2(x^{2}-4x)=5\\\\2[x^{2} - 2(x)(2)]=5](https://tex.z-dn.net/?f=2%28x%5E%7B2%7D-4x%29%3D5%5C%5C%5C%5C2%5Bx%5E%7B2%7D%20-%202%28x%29%282%29%5D%3D5)
The square of difference is written as:
Equation 1
If we compare the given equation from previous step to formula in Equation 1, we note that we have square of first term(x), twice the product of 1st term(x) and second term(2) and the square of second term(2) is missing. So in order to complete the square we need to add and subtract square of 2 to right hand side. i.e.
![2[x^{2}-2(x)(2)+(2)^2-(2)^{2}]=5\\\\ 2[x^{2}-2(x)(2)+(2)^2]-2(2)^{2}=5\\\\ 2(x-2)^{2}-2(4)=5\\\\ 2(x-2)^2-8=5\\\\ 2(x-2)^{2}-8-5=0\\\\ 2(x-2)^{2}-13=0](https://tex.z-dn.net/?f=2%5Bx%5E%7B2%7D-2%28x%29%282%29%2B%282%29%5E2-%282%29%5E%7B2%7D%5D%3D5%5C%5C%5C%5C%202%5Bx%5E%7B2%7D-2%28x%29%282%29%2B%282%29%5E2%5D-2%282%29%5E%7B2%7D%3D5%5C%5C%5C%5C%202%28x-2%29%5E%7B2%7D-2%284%29%3D5%5C%5C%5C%5C%202%28x-2%29%5E2-8%3D5%5C%5C%5C%5C%202%28x-2%29%5E%7B2%7D-8-5%3D0%5C%5C%5C%5C%202%28x-2%29%5E%7B2%7D-13%3D0)
Comparing the above equation with the given equation:
, we can say:
p = 2 and q= -13