Question is in the picture Thanks in Advance
2 answers:
For this question, we can use the same side interior angle theorem.
Angles 1 and 2 are same side interior angles, which means that they are the interior angles on the same side.
To find the angle measure of the angles, we can plug in the value we have.
To find if the oars are parallel, we can use the same side interior angle theorem, which states that if the same side interior angles are supplementary, the lines are parallel.
Seeing as the angles are supplementary, the oars are parallel to each other.
Hope this helps.
頑張って!
You might be interested in
A linear equation can be written as:
y = a*x + b
where a is the slope and b is the y-intercept.
If we know that the line passes through the points (x₁, y₁) and (x₂, y₂) the slope is:
The answers are:
a) y = (-12 L/min)*t + 9,000 L
b) 6,840 L
a) Here we know that we can model this situation with a linear equation, we know that the leak starts at 6:00 am, then we define t = 0 as 6:00 am
We know that at 6:42 am (at t = 42 mins) the tank has 8,496 liters.
And at 7:15 am (at t = 75 mins) the tank has 8,100 liters.
Then we have two points of the line:
(42 mins, 8,496 liters)
(75 mins, 8,100 liters)
Then the slope is:
So the line will be something like:
y = (-12 L/min)*t + b
To find the value of b, which is the initial volume of water in the tank, we use the point (75 mins, 8,100 liters)
This means that if t = 75 mins, then we have y = 8,100 L
So we can replace these two on the above equation:
8,100 L = (-12 L/min)*75 mins + b
8,100L = -900L + b
8,100L + 900L = b = 9,000 L
Then the linear equation is:
y = (-12 L/min)*t + 9,000 L
b) Now we want to know how much water is in the tank at 9:00 am.
Notice that 9:00 am is 3 hours after 6:00 am, our t = 0.
Then 9:00am is equivalent to t = 180 mins.
So to know the volume of water that there is in the tank at 9:00 am, we just need to evaluate the linear function at t = 180 mins.
y = (-12 L/min)*180 mins + 9,000 L = 6,840 L
If you want to learn more, you can read: