CF = CD + DF (since D is simply a point on the line, and C to that point added onto F to that point is CF), so 4y-9+2y-7=14. Adding it up, we get 6y-16=14 and adding by 16 we get 30=6y. Dividing both sides by 6, we get y=5. Plugging that into DF, we get 10-7=DF=3
It factorises to -
4(x+6)
Greetings.
The range is the set of y-value.
The range starts from the minimum point to maximum point.
Our minimum point starts at 0 and maximum point starts less than infinity.
Therefore the range is 0<=y<+inf
However, we do not often write that, although it is right.
Therefore we write as y≥0
Thus, the answer is B choice.
Answer:

Step-by-step explanation:
Given equation can be re written as
............(i)
Now it is given that y(π/2) = 2e
Applying value in (i) we get
ln(2e) = sin(π/2) + c
=> ln(2) + ln(e) = 1+c
=> ln(2) + 1 = 1 + c
=> c = ln(2)
Thus equation (i) becomes
ln(y) = sin(x) + ln(2)
ln(y) - ln(2) = sin(x)
ln(y/2) = sin(x)

Answer:

Step-by-step explanation:
No it isn't, because:
![{1.08}^{ \frac{1}{5} } = { \frac{108}{100} }^{ \frac{1}{5} } = { \frac{27}{25} }^{ \frac{1}{5} } = \sqrt[5]{ \frac{27}{25} } = 1.01551](https://tex.z-dn.net/?f=%20%7B1.08%7D%5E%7B%20%5Cfrac%7B1%7D%7B5%7D%20%7D%20%20%3D%20%20%7B%20%5Cfrac%7B108%7D%7B100%7D%20%7D%5E%7B%20%5Cfrac%7B1%7D%7B5%7D%20%7D%20%20%3D%20%20%7B%20%5Cfrac%7B27%7D%7B25%7D%20%7D%5E%7B%20%5Cfrac%7B1%7D%7B5%7D%20%7D%20%20%3D%20%20%5Csqrt%5B5%5D%7B%20%5Cfrac%7B27%7D%7B25%7D%20%7D%20%20%3D%201.01551)