Yes, it is continuous to its domain
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Explanation:</h2><h2>
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In order to find the domain of the function we need to get the restrictions:
1. From natural log:
2. From quotient:
Matching these two restrictions the domain is:
So the function is continuous to its domain because is defined for every x-value in the interval 
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Answer:
(-4, 2)
Step-by-step explanation:
4x+2y=-12
3x+y=-10
Start by dividing the first equation by 2 to simplify it...
2x+y=-6
Then, subtract -2x from both sides to isolate y...
y=-2x-6
Substitute -2x-6 for y in the second equation...
3x-2x-6=-10
Combine like terms...
x-6=-10
Add 6 to both sides
x-6+6=-10+6
x=-4
Plug -4 in for x to solve for y:
3(-4)+y=-10
-12+y=-10
Add 12 to both sides
-12+12+y=-10+12
y=2
(x,y)=(-4,2)
<u>Answer:</u>
An obtuse triangle has two angles 45 and 18. The value of the unknown angle is 
<u>Solution:</u>
Given that two angles of an obtuse triangle is 45 and 18.
Third angle is θ
We are asked to find the value of third unknown angle theta
According to <em>angle sum property of triangle</em>, sum of three angles of triangle is 
This means in our case sum of 45, 18 and θ should be
Hence value of unknown third angle of an obtuse triangle is
Answer:
The ship is located at (3,5)
Explanation:
In the first test, the equation of the position was:
5x² - y² = 20 ...........> equation I
In the second test, the equation of the position was:
y² - 2x² = 7 ..............> equation II
This equation can be rewritten as:
y² = 2x² + 7 ............> equation III
Since the ship did not move in the duration between the two tests, therefore, the position of the ship is the same in the two tests which means that:
equation I = equation II
To get the position of the ship, we will simply need to solve equation I and equation II simultaneously and get their solution.
Substitute with equation III in equation I to solve for x as follows:
5x²-y² = 20
5x² - (2x²+7) = 20
5x² - 2y² - 7 = 20
3x² = 27
x² = 9
x = <span>± </span>√9
We are given that the ship lies in the first quadrant. This means that both its x and y coordinates are positive. This means that:
x = √9 = 3
Substitute with x in equation III to get y as follows:
y² = 2x² + 7
y² = 2(3)² + 7
y = 18 + 7
y = 25
y = +√25
y = 5
Based on the above, the position of the ship is (3,5).
Hope this helps :)
