Find f(2) if f(x)=2x+1<br> hey besties

Find 2 consecutive odd integers sum of who's square is 290 pls send fast

ELEN [110]

Answer:

<h3>Hence required integers are 11 and 13</h3>

Let x an odd positive integer

Then, according to question

x^2 +(x+2)^2=290

2x^2 +4x−286=0

x^2 +2x−143=0

x ^2+13x−11x−143=0

(x+13)(x−11)=0

x=11 as x is positive

Hence required integers are 11 and 13

Step-by-step explanation:

Hope it is helpful....

7 0

3 years ago

What is the value of w?<br><br><br> A.) 7<br> B.) 3.5<br> C.) 7 square root of 3<br> D.) 14

sammy [17]

D. 14

LL = 1/2 * Hsqrt3
7sqrt3 = 1/2 * ysqrt3
24.25 = ysqrt3
24.25/sqrt3 = y
y = 14

4 0

3 years ago

Read 2 more answers

Which is bigger 3 yards is 10 feet

77julia77 [94]

1 yard = 3 feet.

3 yards = 3 * 3 = 9 feet

9 feet < 10 feet

Hope that helped =)

5 0

3 years ago

Read 2 more answers

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

Describe the relationship between the perimeter of a triangle and the perimeter of the triangle that is formed by the midsegment

lilavasa [31]

It is 1/4 of the area of the larger triangle...basically if you connect the three midpoints of the triangle's sides it creates four identical smaller triangles...

The perimeter is one half of the larger triangle...

3 0

2 years ago

Find f(2) if f(x)=2x+1 hey besties