Identify the type of function represented

Exponential decay.

Step-by-step explanation:

You can use a graphing utility to check this pretty quickly, but you can also look at the equation and get the answer. Since the function has a variable in the exponent, it definitely won't be a linear equation. Quadratic equations are ones of the form ax^2 + bx + c, and your function doesn't look like that, so already you've ruled out two answers.

From the start, since we have a variable in the exponent, we can recognize that it's exponential. Figuring out growth or decay is a little more complicated. Having a negative sign out front can flip the graph; having a negative sign in the exponent flips the graph, too. In your case, you have no negatives; just 2(1/2)^x. What you need to note here, and you could use a few test points to check, is that as x gets bigger, (1/2) will get smaller and smaller. Think about it. When x = 0, 2(1/2)^0 simplifies to just 2. When x = 1, 2(1/2)^1 simplifies to 1. Already, we can tell that this graph is declining, but if you want to make sure, try a really big value for x, like 100. 2(1/2)^100 is a value very very very veeery close to 0. Therefore, you can tell that as the exponent gets larger, the value of the function goes down and gets closer and closer to zero. This means that it can't be exponential growth. In the case of exponential growth, as the exponent gets bigger, your output should increase, too.

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

homozygous. one is homozygous, while two is heterozygous.