200,000+10,000+8,000+300+70+5
Answer:did u find rthe answer please lmk
Step-by-step explanation:
Look at the graph below carefully
Observe the results of shifting ={2}^{x}f(x)=2x
vertically:
The domain, (−∞,∞) remains unchanged.
When the function is shifted up 3 units to ={2}^{x}+3g(x)=2x +3:
The y-intercept shifts up 3 units to (0,4).
The asymptote shifts up 3 units to y=3y=3.
The range becomes (3,∞).
When the function is shifted down 3 units to ={2}^{x}-3h(x)=2 x −3:
The y-intercept shifts down 3 units to (0,−2).
The asymptote also shifts down 3 units to y=-3y=−3.
The range becomes (−3,∞).
Answer:1/5
Step-by-step explanation:
Two and ten divided by two equals 1 and 5 so 1/5
Because the cosine and sine must be negative when evaluated in theta, the angle lies on the third quadrant.
<h3>
In which quadrant is the endpoint of the segment that defines the angle?</h3>
We know that if:
- cos(θ) > 0, then we are on the first or fourth quadrant
- sin(θ) > 0, then we are on first or second quadrant.
Here we know that:
sec(θ) < 0
And we know that:
sec(θ) = 1/cos(θ)
Then we have cos(θ) < 0
We also have that:
sec(θ)*csc(θ) > 0
Because sec(θ) < 0, we must have that csc(θ) < 0.
Remember: csc(θ) = 1/sen(θ)
Then sen(θ) < 0.
Then we have the two conditions:
sen(θ) < 0
cos(θ) < 0
- The cosine is negative on the third and second quadrants.
- The sine is negative on the third and fourth quadrants.
The only quadrant where both are negative is the third quadrant, so that is the correct option.
If you want to learn more about trigonometric functions:
brainly.com/question/8120556
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