Ms.Sanchez drove 96 miles in 1.5 hours

The function g(x)=(x-2^2. The function f(x)=g(x) +3

OlgaM077 [116]

The parent function is:
y = x ^ 2
Applying the following function transformation we have:
Horizontal translations:
Suppose that h> 0
To graph y = f (x-h), move the graph of h units to the right.
We have then:
g (x) = (x-2) ^ 2
Then, we have the following function transformation:
Vertical translations
Suppose that k> 0
To graph y = f (x) + k, move the graph of k units up.
We have then that the original function is:
g (x) = (x-2) ^ 2
Applying the transformation we have
f (x) = g (x) +3
f (x) = (x-2) ^ 2 + 3
Answer:
the function f(x) moves horizontally 2 units rigth.
The function f (x) is shifted vertically 3 units up.

4 0

3 years ago

A pipe is 36 feet long. It needs to be cut into pieces that are each 3/4 feet long. How many pieces can be made from the pipe?

Ket [755]

108/4 = 27 pieces you multiply 36 x 3 = 108 and the denominator stays the same so 108/4 equals to 27 pieces

4 0

3 years ago

Can someone please HELP ME

nirvana33 [79]

a) the independent variable is the number of training miles.

b) the dependent variable is the cost to mail the package

c) the dependent variable is P

3 0

3 years ago

A solution of the equation 3x + y = 15?

Tamiku [17]

Step-by-step explanation:

Since there is 2 unknown variable ,

so we need 2 equation to find their values.

8 0

2 years ago

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

Oksana_A [137]

Answer:

$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

${n \choose 5 } = \frac{n!}{5!(n-5)!}$

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 ${n \choose 4} .$

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is $4 * {n \choose 4} .$

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is ${n \choose 3} .$

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have ${3 \choose 2 } = 3$ possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of $6 * {n \choose 3}$

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of ${n \choose 2}$ possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.

The total amount of possibilities for this case is

$4 * {n \choose 2}$

Case (5): All numbers are the same

This is easy, he have as many possibilities as numbers the set has. In other words, n

Conclussion

By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is

$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

I hope that works for you!

4 0

3 years ago