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2 years ago

Can anyone help me wit this problem..

1 answer:

5 0

Use formula for the number of combinations $\\C_k^n= \frac{n!}{(n-k)!k!} \\ \\n=12, k=2
\\ C_2^{12}= \frac{12!}{(12-2)!2!}= \frac{12!}{10!2!}= \frac{12\times11}{2\times1}=66$

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What is the simplest form of this expression

tekilochka [14]

Answer:

In general, an expression is in simplest form when it is easiest to use. Example, this: 5x + x − 3. Is simpler as: 6x − 3. Common ways to help you simplify: • Combine Like Terms.Step-by-step explanation:sorry if this is isin't right

7 0

2 years ago

HELP !!!!!!!!!!!!!!!!!!

4vir4ik [10]

Answer:

answer is C ( $3^{4}$ )

Step-by-step explanation:

so here is how you do it.

first you need to find out what $3^{8}$ is

which is the same as 3 X 3 X 3 X 3 X 3 X 3 X 3 X 3

the answer to that is 6,561

then you have to square root that

which looks like this

$\sqrt{6,561}$

which is 81

which means that you have to find the same expression that equals to 81

I then did $3^{4}$ (C) to see if the answer will be 81 and it is. so your answer is C

3 0

3 years ago

A school wishes to enclose its rectangular playground using 480 meters of fencing.

Harlamova29_29 [7]

Answer:

Part a) $A(x)=(-x^2+240x)\ m^2$

Part b) The side length x that give the maximum area is 120 meters

Part c) The maximum area is 14,400 square meters

Step-by-step explanation:

The picture of the question in the attached figure

Part a) Find a function that gives the area A(x) of the playground (in square meters) in terms of x

we know that

The perimeter of the rectangular playground is given by

$P=2(L+W)$

we have

$P=480\ m\\L=x\ m$

substitute

$480=2(x+W)$

solve for W

$240=x+W\\W=(240-x)\ m$

Find the area of the rectangular playground

The area is given by

$A=LW$

we have

$L=x\ m\\W=(240-x)\ m$

substitute

$A=x(240-x)\\A=-x^2+240x$

Convert to function notation

$A(x)=(-x^2+240x)\ m^2$

Part b) What side length x gives the maximum area that the playground can have?

we have

$A(x)=-x^2+240x$

This function represent a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the length that give the maximum area that the playground can have

Convert the quadratic equation into vertex form

$A(x)=-x^2+240x$

Factor -1

$A(x)=-(x^2-240x)$

Complete the square

$A(x)=-(x^2-240x+120^2)+120^2$

$A(x)=-(x^2-240x+14,400)+14,400$

$A(x)=-(x-120)^2+14,400$

The vertex is the point (120,14,400)

therefore

The side length x that give the maximum area is 120 meters

Part c) What is the maximum area that the playground can have?

we know that

The y-coordinate of the vertex represent the maximum area

The vertex is the point (120,14,400) -----> see part b)

therefore

The maximum area is 14,400 square meters

Verify

$x=120\ m$

$W=(240-120)=120\ m$

The playground is a square

$A=120^2=14,400\ m^2$

8 0

3 years ago

Find the equation of the line. Use exact numbers.

iren2701 [21]

Answer:

y=3/4x -2

Step-by-step explanation:

I know the last box has a plus sign in front of it but it's still -2 since the y- intercept is at -2. It's the same thing but I just wanted to make sure you knew that in the last box, you still put -2.

Sorry if what I just said above is confusing, i don't know how to explain it better...

4 0

2 years ago

A rectangular field on a farm is to be fenced in using the wall of the barn for one side and 200 meters of fencing for the other

vitfil [10]

Answer:

Rectangular area as a function of x : A(x) = 200*x + 2*x²

A(max) = 5000 m²

Dimensions:

x = 50 m

l = 100 m

Step-by-step explanation:

"x" is the length of the perpendicular side to the wall of the rectangular area to be fenced, and we call "l" the other side (parallel to the wall of the barn) then:

A(r) = x* l and the perimeter of the rectangular shape is

P = 2*x + 2*l but we won´t use any fencing material along the wll of the barn therefore

P = 2*x + l ⇒ 200 = 2*x + l ⇒ l = 200 - 2*x (1)

And the rectangular area as a function of x is:

A(x) = x * ( 200 - 2*x) ⇒ A(x) = 200*x + 2*x²

Taking derivatives on both sides of the equation we get:

A´(x) = 200 - 4*x ⇒ A´= 0

Then 200 - 4*x = 0 ⇒ 4*x = 200 ⇒ x = 50 m

We find the l value, plugging the value of x in equation (1)

l = 200 - 2*x ⇒ l = 200 - 2*50 ⇒ l = 100 m

A(max) = 100*50

A(max) = 5000 m²

3 0

3 years ago