There are two aircraft carriers, a and b, and the carrier a is longer in length the carrier b. the total length of these two carriers is 4178 feet, while the difference of their lengths is only 14 feet.

Step-by-step explanation:

Given , two air craft carriers are A and B . The carrier A is longer than the carrier B in length.

Total length of two carriers =4178 feet

Difference of two carriers = 14 feet

Let

Length of carrier A =x  feet

And

Legth of carrier B =y feet

According to question  

The I equation is

x+y=4178

The II equation is

x-y= 14

Substitution method:  In this method step I :  we  find value of x or y from the equation I

Step II:  put the value of x or y in II equation then we get the value of another  varaiable x or y  

Step : Again put the obatined value of variable x or y  from  step II   in equation I then we get value of other variable x or y .

By  using substitution method ,

we find value of x from equation I

x= 4178-y

Now, put the value of x in equation II then we get

4178-y-y=14

4178-2y=14

4178-14=2y

4164=2y

y= 2082

Again , put the value of y in equation I

2082 +x=4178

x= 4178-2082

x=2096

Hence, the length of carrier A = 2096 feet

The length of carrier B= 2082 feet

g(x) = x² + 2

step-by-step explanation:

it is given that the graphs are the same shape.

the only difference is that the red graph is 2 units higher than the blue graph.

that is to say that the red graph is simply the blue graph that has been moved vertically by 2 units in the positive y direction.

hence

g(x) = f(x) + 2

or

g(x) = x² + 2

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]