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2 years ago

15

a factory makes 5/6 of a box every minute. how many boxes can the factory make in 2/3 of a minute? use a visual fraction model o

r equation to determine the solution

1 answer:

faltersainse [42]2 years ago

6 0

Here you go, hope this helps

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Drag and drop the symbol that correctly represents this graph.

lisov135 [29]

A closed dot mean "greater than or equal to" or "less than or equal to"

An open dot means "less than" or "greater than"

In this case, an open dot would mean

X>-2

6 0

2 years ago

Read 2 more answers

Solve x^2-6x+25 = 0 by completing the square.

Oksana_A [137]

First, subtract both sides by 25:

x^2 - 6x = -25

Because we're completing the square, we subtract both sides by 9:

x^2 - 6x - 9 = -34

(x - 3) = 5.831 (about)

x = 8.831

3 0

2 years ago

What is the equation of this trend line?

solmaris [256]

Equation of this line is K = -2J + 28

<u>Step-by-step explanation:</u>

Step 1:

The equation can be written in slope intercept form which is y = mx + b

Step 2:

Calculate slope of the line, m = (y2 - y1)/(x2 - x1)

⇒ m = (24 - 20)/(2 - 4) = 4/-2) = -2

Step 3:

Find y-intercept of the line

⇒ b = 28

Step 4:

Substitute values in the equation

⇒ K = -2J + 28

5 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

erma4kov [3.2K]

Answer:

$\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

Step-by-step explanation:

Given

5 tuples implies that:

$n = 5$

$(h,i,j,k,m)$ implies that:

$r = 5$

Required

How many 5-tuples of integers $(h, i, j, k,m)$ are there such that $n\ge h\ge i\ge j\ge k\ge m\ge 1$

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed: This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

$^{n}C_r => ^{n + 4}C_5$

$^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}$

$^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}$

Expand the numerator

$^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

<u><em>Solved</em></u>

6 0

2 years ago

The length of a swimming pool is 8 m longer than it's width and the area is 105m²

Art [367]

width= 7m

Step-by-step explanation:

Area= L×W

L= (x+8)m

W=8m

105m²= (x+8)m × (x)m

105m²= x² + 8x

-x² - 8x + 105= 0

(x-7) (x+15) = 0

x= 7 x= -15

x=width = 7m

6 0

2 years ago

Read 2 more answers