All categories

3 years ago

Evaluate n²+5 when n=3

Mathematics

1 answer:

kondor19780726 [428]3 years ago

5 0

Answer:

Step-by-step explanation:

When you sqaure a number you multiply it by itself once so it would be (3x3) + 5. Rmemeber PEMDAS, so you multiply 3x3 first to get 9 and then you add 5 to get 14.

You might be interested in

What is Permutations and Combinations?

marissa [1.9K]

Answer:

See below

Step-by-step explanation:

Permutation is to select an object then arrange it and it cares about the orders while Combination is about only selecting an object without caring the orders.

Permutation can be expressed in math as:

$\displaystyle{_n P _r = \dfrac{n!}{(n-r)!} \ \ \ (n \geq r) }$

where n is a number of total object and r is a number of selected object to arrange. Hence. n cannot be less than r.

Now let's see an example of permutation, suppose we have letter A, B and C. I'd like to know how many ways these words can be arranged:

Since there are 3 letters total and 3 selected letters to arrange then:

$\displaystyle{_3 P _3 = \dfrac{3!}{(3-3)!}}\\\\\displaystyle{_3 P _3 = \dfrac{3 \times 2 \times 1}{0!}}\\\\\displaystyle{_3 P _3 = \dfrac{6}{1}}\\\\\displaystyle{_3 P _3 = 6}$

Therefore, there are 6 ways to arrange the letters - we can also demonstrate visually:

ABC - 1

ACB - 2

BAC - 3

BCA - 4

CAB - 5

CBA - 6

Notice that if you do visually, you'll get the same answer as the calculation of permutation!

----

Combination can be expressed mathematically as:

$\displaystyle{_n C _r = \dfrac{n!}{(n-r)!r!} = \dfrac{_n P _r}{r!} \ \ \ (n \geq r) }$

The difference between permutation and combination is that you only find how many ways you can select object in combination. Therefore, no arrange and doesn't care about order, just ways to select.

Suppose we have same 3 letters: A, B and C. I want to find how many ways I can select these 3 letters:

Since there are 3 letters total and 3 selected letters:

$\displaystyle{_3 C _3 = \dfrac{3!}{(3-3)!3!}}\\\\\displaystyle{_3 C _3 = \dfrac{3!}{0!3!}}\\\\\displaystyle{_3 C _3 = \dfrac{3!}{3!}}\\\\\displaystyle{_3 C _3 = 1}$

Hence, there is only one way to select 3 letters. This makes sense because if you have 3 letters then you can only select 3 letters only one way.

5 0

2 years ago

Suppose F⃗ (x,y)=(x+3)i⃗ +(6y+3)j⃗ . Use the fundamental theorem of line integrals to calculate the following.

Scorpion4ik [409]

In order to use the fundamental theorem of line integrals, you need to find a scalar potential function - that is, a scalar function f(x, y) for which

grad f(x, y) = F(x, y)

This amounts to solving for f such that

∂f/dx = x + 3

∂f/∂y = 6y + 3

Integrating both sides of the first equation with respect to x gives

f = 1/2 x ^2 + 3x + g(y)

Differentiating with respect to y gives

∂f/∂y = dg/dy = 6y + 3

Solving for g gives

g = ∫ (6y + 3) dy = 3y ^2 + 3y + C

and hence

f(x, y) = 1/2 x ^2 + 3x + 3y ^2 + 3y + C

(a) By the fundamental theorem, the integral of F along any path starting at the point P (1, 0) and ending at Q (3, 3) is

∫ F(x, y) • dr = f (3, 3) - f (1, 0) = 99/2 - 7/2 = 46

(b) Now we're talking about a closed path, so the integral is simply 0. We can verify this by checking the integral over the origin-containing paths:

• From the origin to P :

∫ F(x, y) • dr = f (1, 0) - f (0, 0) = 7/2 - 0 = 7/2

• From Q back to the origin:

∫ F(x, y) • dr = f (0, 0) - f (3, 3) = 0 - 99/2 = -99/2

Then the total integral is 7/2 + 46 - 99/2 = 0, as expected.

6 0

2 years ago

(-2.04) (-0.04) What is the product of the expression above?

aleksley [76]

Answer:

0.0816

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

spayn [35]

Answer:81

Step-by-step explanation:

3 0

2 years ago

Need help with this (Venn Diagram with with Exponents, Radicals, and Polynomials)

sertanlavr [38]

Answer:

true, false, true, true

Step-by-step explanation:

The set names in this diagram have nothing to do with exponents, radicals, and polynomials. We'll take the diagram at face value.

(a) The labels on the sets seem to be appropriately placed.

(b) "Some" in this context means "any or all of the set". Since all of the circle representing integers is outside the rectangle representing irrational numbers, it is TRUE that some integer are not irrational numbers.

No part of the circle representing whole numbers is inside the rectangle representing irrational numbers, so it is FALSE that some whole numbers are irrational numbers.

A portion of the circle representing integers is outside the circle representing whole numbers, so it is TRUE that some integers are not whole numbers.

Every part of the circle representing whole numbers is inside the rectangle representing rational numbers, so it is TRUE that all whole numbers are rational numbers.

7 0

3 years ago