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gulaghasi [49]
3 years ago
6

All the boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over. After 60 additional boxe

s arrived and no boxes were removed, all the boxes in the warehouse were arranged in stacks of 14 boxes each, with no boxes left over. How many boxes were in the warehouse before the 60 additional boxes arrived? (1) There were fewer than 110 boxes in the warehouse before the 60 additional boxes arrived. (2) There were fewer than 120 boxes in the warehouse after the 60 additional boxes arrived.
Mathematics
1 answer:
AveGali [126]3 years ago
3 0

Answer:

(1) There were fewer than 110 boxes in the warehouse before the 60 additional boxes arrived. (2) There were fewer than 120 boxes in the warehouse after the 60 additional boxes arrived.

Step-by-step explanation:

Given that all the boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over.

This implies that boxes are in multiples of 12

i.e. 12, or 24, or 36 or....

Let us say this as 12m for m an integer

Now additional boxes arrived =60

Total number of boxes now = 12m+60

This when arranged in stacks of 14,nothing left out

Or this is multiple of 14

12m+60 = 14n for some positive integer n.

By trial and error we find that if 60 is to be divided by 14, 24 should be added and 24 is a multiple of 12.

So we say originally 24 boxes were there now 84 boxes there.

Hence both options are right.

(1) There were fewer than 110 boxes in the warehouse before the 60 additional boxes arrived. (2) There were fewer than 120 boxes in the warehouse after the 60 additional boxes arrived.

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klemol [59]

Answer:

x^2-10x+18=0

<em>a</em> = -10 and <em>b</em> = 18.

Step-by-step explanation:

Let <em>w</em> represent the width of the rectangle.

We are given that the perimeter of the rectangle is 20 cm, this means that:

20=2(x + w)

Let's put <em>w</em> in terms of <em>x</em>. Divide both sides by two:

10=x+w

And solve for <em>w:</em>

<em />w=10-x<em />

So, the rectangle measures <em>x</em> by (10 -<em> x</em>) cm.

<em />

According to the Pythagorean Theorem:

a^2+b^2=c^2

<em>a</em> and <em>b</em> are the legs and <em>c</em> is the hypotenuse.

Substitute <em>x</em> for <em>a, w</em> for <em>b</em>, and 8 for <em>c:</em>

<em />x^2+w^2=8^2<em />

Simplify and substitute:

x^2+(10-x)^2=64

Square:

x^2+(100-20x+x^2)=64

Isolate the equation. So:

2x^2-20x+36=0

Since the leading coefficient is one, divide both sides by two:

x^2-10x+18=0

Therefore, <em>a</em> = -10 and <em>b</em> = 18.

6 0
3 years ago
A flower shop acquired 80 new customers last year. Costs in the marketing and sales areas were the following:
icang [17]

Using proportions, it is found that the Customer Acquisition Cost was of $1,215.

<h3>What is a proportion?</h3>

A proportion is a fraction of a total amount.

In this problem, the customer acquisition cost is the spending in sales divided by the number of customers added.

80 customers were added, considering costs of 1200 + 9000 + 87000 = $97,200, hence:

97200/80 = $1,215.

More can be learned about proportions at brainly.com/question/24372153

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2 years ago
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Yuliya22 [10]

The total bill paid for a person who uses 600 kWh in a month is $60.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Let us assume that the utility company charges $0.10 per kWh, hence:

Total charge in a month = 0.10 * 600 = $60

The total bill paid for a person who uses 600 kWh in a month is $60.

Find out more on equation at: brainly.com/question/2972832

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1 year ago
What is the solution to the equation? Please explain!
Alexus [3.1K]
J: there is no solution
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Please help, I have been trying for a while :(<br><br>question on photo​
Zigmanuir [339]

Answer:

\frac{x-22}{(x+2)(x-4)}

Step-by-step explanation:

multiply the numerator/denominator of the first fraction by (x - 4)

multiply the numerator/denominator of the second fraction by (x + 2)

this ensures that the fractions have a common denominator

\frac{4}{x+2} - \frac{3}{x-4}

= \frac{4(x-4)}{(x+2)(x-4)} - \frac{3(x+2)}{(x+2)(x-4)} ← subtract numerators leaving the common denominator

= \frac{4x-16-3x-6}{(x+2)(x-4)}

=\frac{x-22}{(x+2)(x-4)}

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