I think you meant to say
![\displaystyle \lim_{t\to2}\frac{t^4-6}{2t^2-3t+7}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bt%5Cto2%7D%5Cfrac%7Bt%5E4-6%7D%7B2t%5E2-3t%2B7%7D)
(as opposed to <em>x</em> approaching 2)
Since both the numerator and denominator are continuous at <em>t</em> = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:
![\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bt%5Cto2%7D%20%5Cfrac%7Bt%5E4%20-%206%7D%7B2t%5E2%20-%203t%20%2B%207%7D%20%3D%20%5Cfrac%7B%5Cdisplaystyle%20%5Clim_%7Bt%5Cto2%7D%28t%5E4-6%29%7D%7B%5Cdisplaystyle%20%5Clim_%7Bt%5Cto2%7D%282t%5E2-3t%2B7%29%7D)
Because these expressions are continuous at <em>t</em> = 2, we can compute the limits by evaluating the limands directly at 2:
![\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bt%5Cto2%7D%20%5Cfrac%7Bt%5E4%20-%206%7D%7B2t%5E2%20-%203t%20%2B%207%7D%20%3D%20%5Cfrac%7B%5Cdisplaystyle%20%5Clim_%7Bt%5Cto2%7D%28t%5E4-6%29%7D%7B%5Cdisplaystyle%20%5Clim_%7Bt%5Cto2%7D%282t%5E2-3t%2B7%29%7D%20%3D%20%5Cfrac%7B2%5E4-6%7D%7B2%5Ccdot2%5E2-3%5Ccdot2%2B7%7D%20%3D%20%5Cboxed%7B%5Cfrac%7B10%7D9%7D)
Answer:
Function: ![x+y=9](https://tex.z-dn.net/?f=x%2By%3D9)
Not Function:
and ![x=y^2](https://tex.z-dn.net/?f=x%3Dy%5E2)
Step-by-step explanation:
Given
![x+y=9](https://tex.z-dn.net/?f=x%2By%3D9)
![x^2+y^2=1](https://tex.z-dn.net/?f=x%5E2%2By%5E2%3D1)
![x=y^2](https://tex.z-dn.net/?f=x%3Dy%5E2)
Required
Determine if
is a function of ![x](https://tex.z-dn.net/?f=x)
Solving x+y=9
![x+y=9](https://tex.z-dn.net/?f=x%2By%3D9)
Make y the subject of formula
![y = 9 - x](https://tex.z-dn.net/?f=y%20%3D%209%20-%20x)
<em>Hence; y is a function of x</em>
Solving ![x^2+y^2=1](https://tex.z-dn.net/?f=x%5E2%2By%5E2%3D1)
![x^2+y^2=1](https://tex.z-dn.net/?f=x%5E2%2By%5E2%3D1)
Subtract x² from both sides
![y^2=1 - x^2](https://tex.z-dn.net/?f=y%5E2%3D1%20-%20x%5E2)
Square root of both sides
![y =\± \sqrt{1 - x^2}](https://tex.z-dn.net/?f=y%20%3D%5C%C2%B1%20%5Csqrt%7B1%20-%20x%5E2%7D)
This implies that
or ![y =-\sqrt{1 - x^2}](https://tex.z-dn.net/?f=y%20%3D-%5Csqrt%7B1%20-%20x%5E2%7D)
<em>Because </em>
<em> can be any of those two expressions, it is not a function.</em>
Solving ![x=y^2](https://tex.z-dn.net/?f=x%3Dy%5E2)
![x=y^2](https://tex.z-dn.net/?f=x%3Dy%5E2)
Reorder
![y^2 = x](https://tex.z-dn.net/?f=y%5E2%20%3D%20x)
Take square roots
![y = \±\sqrt{x}](https://tex.z-dn.net/?f=y%20%3D%20%5C%C2%B1%5Csqrt%7Bx%7D)
This implies that
or ![y = -\sqrt{x}](https://tex.z-dn.net/?f=y%20%3D%20-%5Csqrt%7Bx%7D)
<em>Because </em>
<em> can be any of those two expressions, it is not a function.</em>
Counting in 10s is faster than counting in ones because counting in 10s gets you to larger numbers faster because you are going by larger numbers.
X in litres is 67% herbicide
y in litres is 46% herbicide
{x+y=42
{0.67x+0.46y=42*0.55
{67x+46y=2310
{46x+46y=1932
21x=378
x=18 litres
A. She need to use 18L
Answer:
no 5/3 is larger
Step-by-step explanation: