3 over 15 space cross times space 13 over 15 space equals

Simplify Rewrite the expression in the form 5 ^ n. (5/10)/(5/12) =

Amiraneli [1.4K]

Answer:

$5^{-2}$

Step-by-step explanation:

$5^{(10-12)}$

5 0

3 years ago

Given the functions f(x) = 2x -1 and g(x) =

Novosadov [1.4K]

Answer:

Please check the explanation.

Step-by-step explanation:

Given

f(x) = 2x - 1

g(x) = 2 - x

a)

f(x) + g(x) = (2x - 1) + (2 - x)

= 2x -1 + 2 - x

= x + 1

b)

f(x) - g(x) = (2x - 1) - (2 - x)

= 2x - 1 - 2 + x

= 3x - 3

c)

g(-5) - f(-5)

Putting x = -5 in g(x) = 2 - x

g(x) = 2 - x

g(-5) = 2 - (-5) = 2+5 = 7

Putting x = -5 in f(x) = 2x - 1

f(x) = 2x - 1

f(-5) = 2(-5) - 1

= -10 - 1

= -11

Thus,

g(-5) - f(-5) = 7 - (-11) = 7+11 = 18

d)

f(x).g(x) = (2x - 1) (2 - x) = -2x² + 5x - 2

e)

f(g(x)) = f(2-x)

= 2(2-x)-1

= 4-2x-1

= 3-2x

6 0

3 years ago

Which of the following correctly completes the square for the equation

rjkz [21]

Answer:

A. (x + 2)^2 = 31.

Step-by-step explanation:

x^2 + 4x = 27

x^2 + 4x - 27 = 0

A. (x + 2)^2 = 31

x^2 + 4x + 4 - 31 = 0

x^2 + 4x - 27 = 0

This option corresponds to the given equation, so it is correct.

B. (x + 2)^2 = 43

x^2 + 4x + 4 - 43 = 0

x^2 + 4x - 39 = 0

This does not correspond to the equation.

C. (x + 4)^2 = 31

x^2 + 8x + 16 - 31 = 0

x^2 + 8x - 15 = 0

This does not correspond to the equation.

D. (x+4)^2 = 43

x^2 + 8x + 16 - 43 = 0

x^2 + 8x - 27 = 0

This does not correspond to the equation.

So, your answer is A. (x + 2)^2 = 31.

Make sure to SUBMIT and check your PREVIOUS answers to make sure they are right! XD Also, hope this helps!

7 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

3b+(-5)=(2b)³(-3)2+6=?

Flauer [41]

Answer:

b=-1

Step-by-step explanation:

given.3b+(-5)=(2b)³(-3)2+6=?

3b-5=8b-6+6

3b-5=8b

-5=8b-3b

-5=5b

-5/5=5b/5

b=-1

-3-5=-8-6+6=?

-8=-8=-8

7 0

2 years ago