Geraldine is asked to explain the limits on the range of an exponential equation using the function f(x) = 2x. She makes these two statements:

1. As x increases infinitely, the y-values are continually doubled for each single increase in x.
2. As x decreases infinitely, the y-values are continually halved for each single decrease in x.

She concludes that there are no limits within the set of real numbers on the range of this exponential function. Which best explains the accuracy of Geraldine’s statements and her conclusion?

Statement 1 is incorrect because the y-values are increased by 2, not doubled.
Statement 2 is incorrect because the y-values are doubled, not halved.
The conclusion is incorrect because the range is limited to the set of integers.
The conclusion is incorrect because the range is limited to the set of positive real numbers.

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12

The answer would be 484

Ithink it is correct but just ask someone else but i am sure it it

both graphs increase on their domain

neither graph has a y-intercept

step-by-step explanation:

the following points can be visualized easily from the graph

there is a hike in both the graphs so the graphs are increasing on their domain.x-intercept of log x and log 6x are differentthere is a vertical asymptote of both the graphs so there is no y-interceptas both the graphs are increasing so there is no horizontal asymptote