Answer:
a = 8
Step-by-step explanation:
x-1 = 0
x = 1
let x³ + ax² -x -8 be p(x)
hence, p(1) = (1)³ + a(1)² -1 -8 = 0 (since, x-1 is a factor)
= 1 + a - 1-8
= a - 8 = 0
therefore, a = 8
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-6}~,~\stackrel{y_1}{-10})\qquad B(\stackrel{x_2}{x}~,~\stackrel{y_2}{-4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ 10=\sqrt{[x-(-6)]^2+[-4-(-10)]^2}\implies 10=\sqrt{(x+6)^2+(-4+10)^2} \\\\\\ 10^2=(x+6)^2+(6)^2\implies 100=x^2+12x+36+36 \\\\\\ 100=x^2+12x+72\implies 0=x^2+12x-28 \\\\\\ 0=(x+14)(x-2)\implies x= \begin{cases} -14\\ \boxed{2} \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_1%7D%7B-6%7D~%2C~%5Cstackrel%7By_1%7D%7B-10%7D%29%5Cqquad%20B%28%5Cstackrel%7Bx_2%7D%7Bx%7D~%2C~%5Cstackrel%7By_2%7D%7B-4%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%2010%3D%5Csqrt%7B%5Bx-%28-6%29%5D%5E2%2B%5B-4-%28-10%29%5D%5E2%7D%5Cimplies%2010%3D%5Csqrt%7B%28x%2B6%29%5E2%2B%28-4%2B10%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%2010%5E2%3D%28x%2B6%29%5E2%2B%286%29%5E2%5Cimplies%20100%3Dx%5E2%2B12x%2B36%2B36%20%5C%5C%5C%5C%5C%5C%20100%3Dx%5E2%2B12x%2B72%5Cimplies%200%3Dx%5E2%2B12x-28%20%5C%5C%5C%5C%5C%5C%200%3D%28x%2B14%29%28x-2%29%5Cimplies%20x%3D%20%5Cbegin%7Bcases%7D%20-14%5C%5C%20%5Cboxed%7B2%7D%20%5Cend%7Bcases%7D)
because B is on the IV Quadrant, the x-coordinate must be positive.
Answer:
Mean: 46 Median: 45 Mode: 45
Step-by-step explanation:
Mean: 41+45+39+56+48+45+42+34+47+62+35+58 = 552 then, 552/12 = 46
Median: Put numbers in order from least to greatest and cross one out on each side until you get to the middle number. 45 and 45 were the last two numbers to be crossed out, the middle of those would simply be 45!
Mode: 45 appears twice while the rest of the numbers only appeared once.
I hope this helped, good luck! :)
Answer:
The dimensions of the rectangle are two unknowns: The length "l" and the width "w"
The perimeter of a rectangle is found as P = 2*l + 2*w
We also know that the length is 5cm more than twice the width. l = 2*w + 5
These two equations gives us a system of linear equations.
P = 2*l + 2*w
l = 2*w + 5
We can use substitution to replace the "l" in the first equation with 2*w + 5
P = 2*(2*w + 5) + 2*w
P = 4*w + 10 + 2*w
P = 6*w + 10
We know that P = 34cm
34 = 6*w + 10
Subtract 10 from both sides
24 = 6 * w
Divide both sides by 6
4 = w
Now that we know the width, we can find the length by substituting 4 for "w" in the second equation.
l = 2*4 + 5
l = 8 + 5
l = 13
The dimensions of the rectangle is 13cm x 4cm
