Answer:
She ran 6.71km in the evening
Step-by-step explanation:
3km 290m = 3.29km
10 - 3.29 = 6.71km
Answer:
![\frac{{y}^{2}+13y-6}{{(y-1)}^{2}(y+7)}](https://tex.z-dn.net/?f=%5Cfrac%7B%7By%7D%5E%7B2%7D%2B13y-6%7D%7B%7B%28y-1%29%7D%5E%7B2%7D%28y%2B7%29%7D)
Step-by-step explanation:
1) Rewrite
in the form
, where a = y and b = 1.
![\frac{y}{{y}^{2}-2(y)(1)+{1}^{2}}+\frac{6}{{y}^{2}+6y-7}](https://tex.z-dn.net/?f=%5Cfrac%7By%7D%7B%7By%7D%5E%7B2%7D-2%28y%29%281%29%2B%7B1%7D%5E%7B2%7D%7D%2B%5Cfrac%7B6%7D%7B%7By%7D%5E%7B2%7D%2B6y-7%7D)
2) Use Square of Difference:
.
![\frac{y}{{(y-1)}^{2}}+\frac{6}{{y}^{2}+6y-7}](https://tex.z-dn.net/?f=%5Cfrac%7By%7D%7B%7B%28y-1%29%7D%5E%7B2%7D%7D%2B%5Cfrac%7B6%7D%7B%7By%7D%5E%7B2%7D%2B6y-7%7D)
3) Factor
.
1 - Ask: Which two numbers add up to 6 and multiply to -7?
-1 and 7
2 - Rewrite the expression using the above.
![(y-1)(y-7)](https://tex.z-dn.net/?f=%28y-1%29%28y-7%29)
Outcome/Result: ![\frac{y}{(y-1)^2} +\frac{6}{(y-1)(y+7)}](https://tex.z-dn.net/?f=%5Cfrac%7By%7D%7B%28y-1%29%5E2%7D%20%2B%5Cfrac%7B6%7D%7B%28y-1%29%28y%2B7%29%7D)
4) Rewrite the expression with a common denominator.
![\frac{y(y+7)+6(y-1)}{{(y-1)}^{2}(y+7)}](https://tex.z-dn.net/?f=%5Cfrac%7By%28y%2B7%29%2B6%28y-1%29%7D%7B%7B%28y-1%29%7D%5E%7B2%7D%28y%2B7%29%7D)
5) Expand.
![\frac{{y}^{2}+7y+6y-6}{{(y-1)}^{2}(y+7)}](https://tex.z-dn.net/?f=%5Cfrac%7B%7By%7D%5E%7B2%7D%2B7y%2B6y-6%7D%7B%7B%28y-1%29%7D%5E%7B2%7D%28y%2B7%29%7D)
6) Collect like terms.
![\frac{{y}^{2}+(7y+6y)-6}{{(y-1)}^{2}(y+7)}](https://tex.z-dn.net/?f=%5Cfrac%7B%7By%7D%5E%7B2%7D%2B%287y%2B6y%29-6%7D%7B%7B%28y-1%29%7D%5E%7B2%7D%28y%2B7%29%7D)
7) Simplify
to ![{y}^{2}+13y-6y](https://tex.z-dn.net/?f=%7By%7D%5E%7B2%7D%2B13y-6y)
![\frac{{y}^{2}+13y-6}{{(y-1)}^{2}(y+7)}](https://tex.z-dn.net/?f=%5Cfrac%7B%7By%7D%5E%7B2%7D%2B13y-6%7D%7B%7B%28y-1%29%7D%5E%7B2%7D%28y%2B7%29%7D)
Answer:
Check below, please
Step-by-step explanation:
Step-by-step explanation:
1.For which values of x is f '(x) zero? (Enter your answers as a comma-separated list.)
When the derivative of a function is equal to zero, then it occurs when we have either a local minimum or a local maximum point. So for our x-coordinates we can say
![f'(x)=0\: at \:x=2, and\: x=-2](https://tex.z-dn.net/?f=f%27%28x%29%3D0%5C%3A%20at%20%5C%3Ax%3D2%2C%20and%5C%3A%20x%3D-2)
2. For which values of x is f '(x) positive?
Whenever we have
![f'(x)>0](https://tex.z-dn.net/?f=f%27%28x%29%3E0)
then function is increasing. Since if we could start tracing tangent lines over that graph, those tangent lines would point up.
![f'(x)>0 \:at [-4,-2) \:and\:(2, \infty)](https://tex.z-dn.net/?f=f%27%28x%29%3E0%20%5C%3Aat%20%5B-4%2C-2%29%20%5C%3Aand%5C%3A%282%2C%20%5Cinfty%29)
3. For which values of x is f '(x) negative?
On the other hand, every time the function is decreasing its derivative would be negative. The opposite case of the previous explanation. So
![f'(x)](https://tex.z-dn.net/?f=f%27%28x%29%20%3C0%20%5C%3A%20at%5C%3A%20%5B-2%2C2%5D)
4.What do these values mean?
![f(x) \:is \:increasing\:when\:f'(x) >0\\\\f(x)\:is\:decreasing\:when f'(x)](https://tex.z-dn.net/?f=f%28x%29%20%5C%3Ais%20%5C%3Aincreasing%5C%3Awhen%5C%3Af%27%28x%29%20%3E0%5C%5C%5C%5Cf%28x%29%5C%3Ais%5C%3Adecreasing%5C%3Awhen%20f%27%28x%29%3C0)
5.(b) For which values of x is f ''(x) zero?
In its inflection points, i.e. when the concavity of the curve changes. Since the function was not provided. There's no way to be precise, but roughly
at x=-4 and x=4
The constant is 3 because it has no variable.
The coefficient is 7b because it has a variable.
Answer:
36 cups of flour
Step-by-step explanation:
if 3 cups of flour=8 ounces.what about 96 onces you cross multiply