I can’t answer this with out the distance they jumped.
Answer:
![\huge\boxed{(f-g)\left(-\dfrac{1}{3}\right)=-3\dfrac{1}{3}}](https://tex.z-dn.net/?f=%5Chuge%5Cboxed%7B%28f-g%29%5Cleft%28-%5Cdfrac%7B1%7D%7B3%7D%5Cright%29%3D-3%5Cdfrac%7B1%7D%7B3%7D%7D)
Step-by-step explanation:
![\text{We have}:\\\\f(x)=x^2-4\ \text{and}\ g(x)=x^2+2x\\\\(f-g)(x)=f(x)-g(x)\\\\\text{therefore}\\\\(f-g)(x)=(x^2-4)-(x^2+2x)=x^2-4-x^2-2x\\\\=(x^2-x^2)-2x-4=-2x-4\\\\(f-g)\left(-\dfrac{1}{3}\right)\to\text{put}\ x=-\dfrac{1}{3}\ \text{to the equation of the function}\ (f-g)(x):\\\\(f-g)\left(-\dfrac{1}{3}\right)=-2\left(-\dfrac{1}{3}\right)-4=\dfrac{2}{3}-4=-3\dfrac{1}{3}](https://tex.z-dn.net/?f=%5Ctext%7BWe%20have%7D%3A%5C%5C%5C%5Cf%28x%29%3Dx%5E2-4%5C%20%5Ctext%7Band%7D%5C%20g%28x%29%3Dx%5E2%2B2x%5C%5C%5C%5C%28f-g%29%28x%29%3Df%28x%29-g%28x%29%5C%5C%5C%5C%5Ctext%7Btherefore%7D%5C%5C%5C%5C%28f-g%29%28x%29%3D%28x%5E2-4%29-%28x%5E2%2B2x%29%3Dx%5E2-4-x%5E2-2x%5C%5C%5C%5C%3D%28x%5E2-x%5E2%29-2x-4%3D-2x-4%5C%5C%5C%5C%28f-g%29%5Cleft%28-%5Cdfrac%7B1%7D%7B3%7D%5Cright%29%5Cto%5Ctext%7Bput%7D%5C%20x%3D-%5Cdfrac%7B1%7D%7B3%7D%5C%20%5Ctext%7Bto%20the%20equation%20of%20the%20function%7D%5C%20%28f-g%29%28x%29%3A%5C%5C%5C%5C%28f-g%29%5Cleft%28-%5Cdfrac%7B1%7D%7B3%7D%5Cright%29%3D-2%5Cleft%28-%5Cdfrac%7B1%7D%7B3%7D%5Cright%29-4%3D%5Cdfrac%7B2%7D%7B3%7D-4%3D-3%5Cdfrac%7B1%7D%7B3%7D)
Answer:
1
Step-by-step explanation:
First find f(0) and g(0). These are the values where x=0 in each function.
f(0) = 1+0 = 1
g(0) = 1^2 - 1 = 1-1 = 0
So f(0) = 1 and g(0) = 0.
Now substitute f(0) = 1 into g(t).
g(1) = 1^2 -1 = 1-1 = 0.
So g(f(0)) = 0.
Now substitute g(0) = 0 into f(t).
f(0) = 1 + 0 = 1.
So f(g(0)) = 1.
Add the values 0 and 1 to get 0+1 = 1.
Answer:
We use scientific notation as sometimes numbers are so so big, that it becomes inconvenient for us to write the actual number.
Step-by-step explanation:
For example, if you have 9 billion of something, you wouldn't want to write those 9 zeroes all the time.
When writing in scientific notation we count the number of decimal places to when we have one non-zero integer to the right of the decimal place. We then write that number and multiply it by ten to the number of decimal places you counted earlier.
e.g. 4.8 X 10^7