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3 years ago

Freeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee pointssssssssssss

Mathematics

2 answers:

VikaD [51]3 years ago

8 0

Answer:

say less hahhahh

Step-by-step explanation:

hahahaha tjx

morpeh [17]3 years ago

4 0

Answer:

tysm!!

Step-by-step explanation:

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Write the slope-intercept form of the equation (-3,3) parallel to y=-5/3x+3

bazaltina [42]

y=5/3x+ 8

The step are:

using the y-y1=m(x-y2)

y-3=5/3(x+3)

y=5/3x+8

5 0

3 years ago

Let A be a finite, non-empty subset of R. Prove that A has a maximum and a minimum. (Recall that a maximum of a set A is an uppe

julia-pushkina [17]

Answer:

Let us use mathematical induction to prove the statement. So, we are going to start checking the statement for the first natural numbers.

$n=1$ : Our set is $\{x_1\}$ . So, obviously, $x_1$ is the maximum and minimum of our set. Then, the statement is true for $n=1$ .

$n=2$ : Our set is $\{x_1,x_2\}$ . Necessarily, $x_1$ or $x_1>x_2$ . In both cases, there is a minimum and a maximum.

Once we have our statement checked for the initial cases, we state our <em>induction hypothesis</em>:

For every finite set A of $n$ elements there exists a maximum and a minimum.

Now, let us prove the that the above assertion is true for sets with $n+1$ elements.

Our set is $A=\{x_1,x_2,\ldots,x_n,x_{n+1}\}$ and we want to find

$\max\{x_1,x_2,\ldots,x_n,x_{n+1}\}$ .

Notice that this problem is equivalent to solve

$\max\{\max\{x_1,x_2,\ldots,x_n\},x_{n+1}\}$ ,

i.e, to find the maximum among $n+1$ numbers, we can find first the miximum among $n$ and then compare with the other one.

Now, using our induction hypothesis we know that there is a maximum in the set $\{x_1,x_2,\ldots,x_n\}$ , because it has $n$ elements. Let us write

$x' =\max\{x_1,x_2,\ldots,x_n\}$ .

So, in order to find the maximum of A, we have to find the maximum of $\A'={x',x_{n+1}\}$ . As we have checked at the beginning, there is a maximum in $A'$ , and it is the maximum of A.

Hence, we have completed the prove for the existence of the maximum of a set with $n+1$ elements. The prove for the existence of the minimum is analogue, we just need to change ‘‘maximum’’ for ‘‘minimum’’.

5 0

3 years ago

Multiplying binomials help please (10 points)

algol [13]

This is the answer 6x^2+15x+9

5 0

3 years ago

Read 2 more answers

The equations y+x^2+6x+8 and y =(x+2) (x+4) both define the same quadratic function. Without graphing, identify the x- and y-int

Goryan [66]

Answer:

we get the