As per the problem,
Rhonda bought a new laptop for $800.
The laptop depreciates, or loses, 20% of its value each year.
The value of the laptop at a later time can be found using the formula
Here we have
P=$800
r=20%=0.20
t=2 years
Substitute the values in equation (1) we get
The laptop be worth in two years will be $512.
Answer:
600 feet
Step-by-step explanation:
Answer:
The correct option is A. x – 1 < n < 3x + 5
Step-by-step explanation:
In a triangle sum of any two sides is always greater than the third side.
Now, the sides of the triangle are given to be :
2x + 2, x + 3 , n
Now, first take 2x + 2 and x + 3 as two sides and the side of length n as third side.
By using the property that sum of two sides is always greater than the third side in a triangle.
⇒ 2x + 2 + x + 3 > n
⇒ 3x + 5 > n ......(1)
Now, take n and x + 3 as two sides and the side of length 2x + 2 as the third side of triangle.
So, by the property, we have :
n + x + 3 > 2x + 2
⇒ n > x - 1 ...........(2)
From both the equations (1) and (2) , We get :
x – 1 < n < 3x + 5
Therefore, The correct option is A. x – 1 < n < 3x + 5
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20N%28%5Cstackrel%7Bx_1%7D%7B-3%7D~%2C~%5Cstackrel%7By_1%7D%7B10%7D%29%5Cqquad%20A%28%5Cstackrel%7Bx_2%7D%7B6%7D~%2C~%5Cstackrel%7By_2%7D%7B3%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20NA%3D%5Csqrt%7B%286%2B3%29%5E2%2B%283-10%29%5E2%7D%5Cimplies%20NA%3D%5Csqrt%7B130%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_2%7D%7B6%7D~%2C~%5Cstackrel%7By_2%7D%7B3%7D%29%5Cqquad%20D%28%5Cstackrel%7Bx_1%7D%7B6%7D~%2C~%5Cstackrel%7By_1%7D%7B-1%7D%29%20%5C%5C%5C%5C%5C%5C%20AD%3D%5Csqrt%7B%286-6%29%5E2%2B%28-1-3%29%5E2%7D%5Cimplies%20AD%3D4%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
now that we know how long each one is, let's plug those in Heron's Area formula.
![\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7BHeron%27s%20area%20formula%7D%20%5C%5C%5C%5C%20A%3D%5Csqrt%7Bs%28s-a%29%28s-b%29%28s-c%29%7D%5Cqquad%20%5Cbegin%7Bcases%7D%20s%3D%5Cfrac%7Ba%2Bb%2Bc%7D%7B2%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D%5Csqrt%7B130%7D%5C%5C%20b%3D4%5C%5C%20c%3D%5Csqrt%7B202%7D%5C%5C%5B1em%5D%20s%3D%5Cfrac%7B%5Csqrt%7B130%7D%2B4%2B%5Csqrt%7B202%7D%7D%7B2%7D%5C%5C%5B1em%5D%20s%5Capprox%2014.81%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D%5Csqrt%7B14.81%2814.81-%5Csqrt%7B130%7D%29%2814.81-4%29%2814.81-%5Csqrt%7B202%7D%29%7D%20%5C%5C%5C%5C%5C%5C%20A%3D%5Csqrt%7B324%7D%5Cimplies%20A%3D18)
(x^4 - 9)(x^3 + 9)
This is a binomial * binomial. Use the FOIL method for multiplying.
First, outer, inner, last.
x^4 * x^3 = x^7
x°^4 * 9 = 9x^4
-9 * x^3 = -9 x^3
-9 * 9 = -81
Putting it back together.
x^7 +9x^4 -9x^3 - 81
