![\bf \lim\limits_{x\to \infty}~\left( \cfrac{1}{8} \right)^x\implies \lim\limits_{x\to \infty}~\cfrac{1^x}{8^x}\\\\[-0.35em] ~\dotfill\\\\ \stackrel{x = 10}{\cfrac{1^{10}}{8^{10}}}\implies \cfrac{1}{8^{10}}~~,~~ \stackrel{x = 1000}{\cfrac{1^{1000}}{8^{1000}}}\implies \cfrac{1}{8^{1000}}~~,~~ \stackrel{x = 100000000}{\cfrac{1^{100000000}}{8^{100000000}}}\implies \cfrac{1}{8^{100000000}}~~,~~ ...](https://tex.z-dn.net/?f=%5Cbf%20%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7D~%5Cleft%28%20%5Ccfrac%7B1%7D%7B8%7D%20%5Cright%29%5Ex%5Cimplies%20%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7D~%5Ccfrac%7B1%5Ex%7D%7B8%5Ex%7D%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7Bx%20%3D%2010%7D%7B%5Ccfrac%7B1%5E%7B10%7D%7D%7B8%5E%7B10%7D%7D%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B8%5E%7B10%7D%7D~~%2C~~%20%5Cstackrel%7Bx%20%3D%201000%7D%7B%5Ccfrac%7B1%5E%7B1000%7D%7D%7B8%5E%7B1000%7D%7D%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B8%5E%7B1000%7D%7D~~%2C~~%20%5Cstackrel%7Bx%20%3D%20100000000%7D%7B%5Ccfrac%7B1%5E%7B100000000%7D%7D%7B8%5E%7B100000000%7D%7D%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B8%5E%7B100000000%7D%7D~~%2C~~%20...)
now, if we look at the values as "x" races fast towards ∞, we can as you see above, use the values of 10, 1000, 100000000 and so on, as the value above oddly enough remains at 1, it could have been smaller but it's constantly 1 in this case, the value at the bottom is ever becoming a larger and larger denominator.
let's recall that the larger the denominator, the smaller the fraction, so the expression is ever going towards a tiny and tinier and really tinier fraction, a fraction that is ever approaching 0.
Answer:
i dont understand
Step-by-step explanation:
Answer:
-1/7
Step-by-step explanation:
The first step is to isolate the variable, x. To do so, add x to both sides. This gives you the new equation: 2 + 7x = 1.
Now, subtract 2 from both sides to further isolate the variable, giving you 7x = -1.
Divide both sides by 7, giving you the final equation of x = -1/7.
First, let's put the second equation, <span>x-2.23y+10.34=0, in terms of y:
x - 2.23y +10.34 = 0
2.23y = x + 10.34
y = .45x + 4.64
Now we can substitute the right side of this equation for y in the first equation
</span><span>y=2x^2+8x
.45x + 4.64 = 2x^2 + 8x
Turn it into a quadratic by getting 0 on one side:
2x^2 + 8x - .45x - 4.64 = 0
2x^2 + 7.55x - 4.64 = 0 Divide both sides by 2
x^2 + 3.76x - 2.32 = 0
x =( -b +/- </span>√(b² - 4ac) ) / 2a
x =( -3.76 +/- √(14.14 + 9.28)) ÷ 2
x = .54 or -4.31
Plug the x values into y = .45x + 4.64
y = .45 (.54) + 4.64
y = 4.88 when x= .54
y = .45 (-4.31) + 4.64
y = 2.70 when x= -4.31
Solution set:
{ (0.54, 4.88) , (-4.31, 2.70) }
I dont know i am just guessing so yeah just give me a l
