Answer:
50%
Step-by-step explanation:
Let the total number of trees be T and suppose that this is made up of eucalyptus trees E and other kinds of trees, O.
Then E+O=T, and since 99% of the trees are eucalyptus trees, EE+O=0.99.
Therefore
0.99E+0.99O=E
0.01E=0.99O
E=99O.
Suppose we reduce the number of eucalyptus trees by some amount x and that the remaining number of eucalyptus trees are now 98% of the total trees.
Then we would have E−x(E−x)+O=0.98.
Therefore,
0.02(E−x)=0.98O
0.02E−0.02x=0.98O
0.02x=0.02E−0.98O
x=E−50(0.98)O
x=E−49O
x=E−49(E99)
x=5099E≈0.505050...×E.
So we end up removing a little over half of the eucalyptus trees. EDIT: Forgot to mention - removing 50.505050...% of the eucalyptus trees means that we are removing 0.50505050....×0.99T which is 0.5T or half the trees in the neighborhood.
Y=x².
This is te function of a parabola open upward
the domain is all values of x ={x/x∈z}
Answer:
5 cm
Step-by-step explanation:
We khow that the altitude of this triangle is 1cm shorter than the base
- Let H be our altitude and B our base and A the area of the triangle
- A= (B*H)/2 ⇒ 15=(B*H)/2
- H is 1cm shorter than B ⇒ B=H+1
- H*(H+1)/2=15 ⇒ H*(H+1)=30⇒ H²+H=30⇒H²+H-30+0
that's a quadratic equation . Let's calculate the dicriminant .
Let Δ be the dicriminant
- a=1
- b=1
- c= -30
- Δ=b²-4*a*c = 1²-4*1*(-30)=1+4*30=121≥0
- Δ≥0⇔ that we have two solutions x and y
- x= (-1-
)/2= (-1-11)/2= -6 - y= (-1+
)/2= 10/2 = 5
We have a negative value and a positive one
The altitude is a distance so it can't be negative
H= 5cm
Let x be the distance traveled on the highway and y the distance traveled in the city, so:
Now, the system of equations in matrix form will be:
![\left[\begin{array}{ccc}1&1&\\ \frac{1}{65} & \frac{1}{25} &\end{array}\right] \left[\begin{array}{ccc}x&\\y&\end{array}\right] = \left[\begin{array}{ccc}375&\\7&\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%26%5C%5C%20%5Cfrac%7B1%7D%7B65%7D%20%26%20%5Cfrac%7B1%7D%7B25%7D%20%26%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%26%5C%5Cy%26%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D375%26%5C%5C7%26%5Cend%7Barray%7D%5Cright%5D%20)
Next, we are going to find the determinant:
![D= \left[\begin{array}{ccc}1&1\\ \frac{1}{65} & \frac{1}{25} \end{array}\right] =(1)( \frac{1}{25}) - (1)( \frac{1}{65} )= \frac{8}{325}](https://tex.z-dn.net/?f=D%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C%20%5Cfrac%7B1%7D%7B65%7D%20%26%20%5Cfrac%7B1%7D%7B25%7D%20%5Cend%7Barray%7D%5Cright%5D%20%3D%281%29%28%20%5Cfrac%7B1%7D%7B25%7D%29%20-%20%281%29%28%20%5Cfrac%7B1%7D%7B65%7D%20%29%3D%20%5Cfrac%7B8%7D%7B325%7D%20)
Next, we are going to find the determinant of x:
![D_{x} = \left[\begin{array}{ccc}375&1\\7& \frac{1}{25} \end{array}\right] = (375)( \frac{1}{25} )-(1)(7)=8](https://tex.z-dn.net/?f=%20D_%7Bx%7D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D375%261%5C%5C7%26%20%5Cfrac%7B1%7D%7B25%7D%20%5Cend%7Barray%7D%5Cright%5D%20%3D%20%28375%29%28%20%5Cfrac%7B1%7D%7B25%7D%20%29-%281%29%287%29%3D8)
Now, we can find x:

Now that we know the value of x, we can find y:

Remember that time equals distance over velocity; therefore, the time on the highway will be:

An the time on the city will be:

We can conclude that the bus was five hours on the highway and two hours in the city.
Answer:
Check out the image below!
Step-by-step explanation: