The domain is all real numbers. (-infinity, infinity)
You can put any real number x value into the function.
Answer:
3x - y = 0; 2x - y = -5
Step-by-step explanation:
Let x be the present age of Hans and y be the present age of Grace,
Since, in present Grace is three times as old as Hans,
⇒ y = 3x
⇒ 3x - y = 0
Now, after 5 years,
The age of Hans = x + 5,
And, the age of Grace = y + 5
Also, in 5 years Grace will be twice as old as Hans is then,
⇒ y + 5 = 2 ( x + 5 )
⇒ y + 5 = 2x + 10
⇒ 2x - y = -5
Hence, the required system of linear equations is,
3x - y = 0; 2x - y = -5
Answer is in the file below
tinyurl.com/wpazsebu
The equation which is equivalent to the equation as given in the task content is; 6x -8y = 36.
<h3>Which equation is equivalent to the equation as given in the task content?</h3>
According to the task content, it follows that the equation which is given in the task content is;
3x -4y = 18.
Hence, upon multiplication of the whole equation by 2; the resulting equation from the multiplication is;
6x -8y = 36
Ultimately, the equivalent equation is; 6x -8y = 36.
Read more on equivalent equations;
brainly.com/question/26437692
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Answer:

Step-by-step explanation:
We want to find the Riemann sum for
with n = 6, using left endpoints.
The Left Riemann Sum uses the left endpoints of a sub-interval:

where
.
Step 1: Find 
We have that 
Therefore, 
Step 2: Divide the interval
into n = 6 sub-intervals of length 
![a=\left[0, \frac{\pi}{8}\right], \left[\frac{\pi}{8}, \frac{\pi}{4}\right], \left[\frac{\pi}{4}, \frac{3 \pi}{8}\right], \left[\frac{3 \pi}{8}, \frac{\pi}{2}\right], \left[\frac{\pi}{2}, \frac{5 \pi}{8}\right], \left[\frac{5 \pi}{8}, \frac{3 \pi}{4}\right]=b](https://tex.z-dn.net/?f=a%3D%5Cleft%5B0%2C%20%5Cfrac%7B%5Cpi%7D%7B8%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B%5Cpi%7D%7B4%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B4%7D%2C%20%5Cfrac%7B3%20%5Cpi%7D%7B8%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B3%20%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%20%5Cfrac%7B5%20%5Cpi%7D%7B8%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B5%20%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B3%20%5Cpi%7D%7B4%7D%5Cright%5D%3Db)
Step 3: Evaluate the function at the left endpoints






Step 4: Apply the Left Riemann Sum formula

