Both of your answers are correct. In the first equation, if you were to substitute 3 for x, because you have to do exponents first, you have 3^2, which is 9, and 9*6 =54. If you were to substitute -3 for x again you would have to do the exponents first which would be -3^2 which is 9, and 9*6 = 54.
For the second question, if I substitute 0.75 for t, I get -9 + 18 + 7. This is equal to 16, and therefore the answer would be D. Hope this helps! :D. Let me know if you have questions about my explanation, and feel free to post more questions.
Answer:
? = 4 units
Step-by-step explanation:
'2' is to ' 7+2 ' as ? is to '14 + ? '
2/ 9 = ? / (14 + ?) cross mutiply
28 + 2? = 9?
28 = 7?
? = 4
It looks like you have the domain confused for the range! You can think of the domain as the set of all "inputs" for a function (all of the x values which are allowed). In the given function, we have no explicit restrictions on the domain, and no situations like division by 0 or taking the square root of a negative number that would otherwise put limits on it, so our domain would simply be the set of all real numbers, R. Inequality notation doesn't really use ∞, so you could just put an R to represent the set. In set notation, we'd write

and in interval notation,

The <em>range</em>, on the other hand, is the set of all possible <em>outputs</em> of a function - here, it's the set of all values f(x) can be. In the case of quadratic equations (equations with an x² term), there will always be some minimum or maximum value limiting the range. Here, we see on the graph that the maximum value for f(x) is 3. The range of the function then includes all values less than or equal to 3. As in inequality, we can say that
,
in set notation:

(this just means "f(x) is a real number less than or equal to 3")
and in interval notation:
![(-\infty,3]](https://tex.z-dn.net/?f=%20%28-%5Cinfty%2C3%5D%20)
Answer:
10/9
Step-by-step explanation:
Answer: After 7 years the population will be one-half the initial amount.
Step-by-step explanation:
Given: Initial population = 500,000
The population declines according to the equation:
, where P is the population in t years later.
One-half the initial amount = 0.5 x 500,000
= 250,000
Put P(t)=250,000, we get

Hence, After 7 years the population will be one-half the initial amount.