<span>The correct answer is option B. i.e. Both equation 1 and 2 have the same number of solutions. The equation 1 is |5x+6| = 41 By removing the modulus sign we get, 5x + 6 = 41 and 5x +6 = -41. Solving, 5x = 6 = 41 we get, 5x = 35 or x = 7. And, on solving 5x +6 = -41 we get, 5x = -47 or x = 9.4 Now. the second equation is |2x+13| = 28 By removing the modulus sign we get, 2x+ 13 = 28 and 2x +13 = -28. Solving, 2x = 28 -13 = 15 or x = 7.5 . And, on solving 2x +13 = -28 we get, 2x = -41 or x = -20.5.TH </span>
Analysis:
1) The graph of function f(x) = √x is on the first quadrant, because the domain is x ≥ 0 and the range is y ≥ 0
2) The first transformation, i.e. the reflection of f(x) over the x axis, leaves the function on the fourth quadrant, because the new image is y = - √x.
3) The second transformation, i.e. the reflection of y = - √x over the y-axis, leaves the function on the third quadrant, because the final image is - √(-x). This is, g(x) = - √(-x).
From that you have, for g(x):
* Domain: negative x-axis ( -x ≥ 0 => x ≤ 0)
* Range: negative y-axis ( - √(-x) ≤ 0 or y ≤ 0).
Answers:
Now let's examine the statements:
<span>A)The functions have the same range:FALSE the range changed from y ≥ 0 to y ≤ 0
B)The functions have the same domains. FALSE the doman changed from x ≥ 0 to x ≤ 0
C)The only value that is in the domains of both functions is 0. TRUE: the intersection of x ≥ 0 with x ≤ 0 is 0.
D)There are no values that are in the ranges of both functions. FALSE: 0 is in the ranges of both functions.
E)The domain of g(x) is all values greater than or equal to 0. FALSE: it was proved that the domain of g(x) is all values less than or equal to 0.
F)The range of g(x) is all values less than or equal to 0.
TRUE: it was proved above.</span>
Answer:
60
Step-by-step explanation:
This is a simple question.
Good luck!
The answer will either be E or 38
Answer: B Because if u look at the question u should be able to see that its b im not giving hints but its b
