Andy is solving a quadratic equation using completing the square. If a step in the process results in StartFraction 169 Over 9 E

The answer is A. 55. Explanation: For the 35%, take your 200 and divide it by 2 to yield two 100's. Since 35% equates to a portion of 100%, grab 35 from each 100 and sum them up to reach 70. For the proportion of 3/8, divide which results in 0.375 then multiply by 200 to ascertain that 75 are yellow tags. Adding 75 to 70 provides a total of 145, which when subtracted from 200 results in 55, indicating there are 55 red tags.

The formula to calculate the difference between two standard deviations of populations n1 and n2 is:
sigma (difference)=√(sigma1/n1 + sigma2/n2). For this scenario:

sigma(d)= √(49/100 + 36/50)

Thus, the calculated standard deviation of the difference equals 1.1.

Response:

Explanatory steps:

I am unsure as well. However, I can state that the first one is

Sarah is 40 years old and her mother is 64 years old.

Answer: We start with equation 1

x=64/3

Next, we solve equation 2

y=125/3

Now, we combine equations 1 and 2

y-x=125/3 -64/3

=61/3

Step-by-step explanation:

Answer:

We have defined functions:

f(x) = IxI + 1

g(x) = 1/x^3.

Currently, it is evident that the composite functions are not commutative.

How can we demonstrate this?

To determine if two composite functions are commutative, the following must hold true:

f(g(x)) = g(f(x))

One could apply brute force (simply substituting values to see if the composite functions commute),

but I will opt for a more sophisticated approach.

There are two notable observations:

g(x) has a point of discontinuity at x = 0.

Thus:

f(g(x)) = I 1/x^3 I + 1

remains discontinuous at x = 0, whereas:

g(f(x)) = 1/(IxI + 1)^3

shows that the denominator IxI + 1 can never reach zero.

At this point, there is no discontinuity.

Consequently, the composite functions cannot be commutative.