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Semmy [17]
3 years ago
12

On the day a coffee shop first opened it had 140 customers 65% of the customers ordered a large coffee how manny customers order

ed a large coffee?
Mathematics
2 answers:
madam [21]3 years ago
8 0
91 patrons of the coffee shop ordered a large

Illusion [34]3 years ago
4 0

Answer:

140/100 * 65 = 91 people

Divide 140 by 100 to find 1 percent then multiply by 65 to find 65 percent.

Step-by-step explanation:

n/a

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The perimeter of a regular hexagon is 78 cm .What will be the length of each side?
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Question Answer

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3 years ago
A rectangular parking lot has an area of 15,000 feet squared, the length is 20 feet more than the width. Find the dimensions
faust18 [17]

Dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet

<h3><u>Solution:</u></h3>

Given that  

Area of rectangular parking lot = 15000 square feet

Length is 20 feet more than the width.

Need to find the dimensions of rectangular parking lot.

Let assume width of the rectangular parking lot in feet be represented by variable "x"

As Length is 20 feet more than the width,

so length of rectangular parking plot = 20 + width of the rectangular parking plot

=> length of rectangular parking plot = 20 + x = x + 20

<em><u>The area of rectangle is given as:</u></em>

\text {Area of rectangle }=length \times width

Area of rectangular parking lot = length of rectangular parking plot \times width of the rectangular parking

\begin{array}{l}{=(x+20) \times (x)} \\\\ {\Rightarrow \text { Area of rectangular parking lot }=x^{2}+20 x}\end{array}

But it is given that Area of rectangular parking lot = 15000 square feet

\begin{array}{l}{=>x^{2}+20 x=15000} \\\\ {=>x^{2}+20 x-15000=0}\end{array}

Solving the above quadratic equation using quadratic formula

<em><u>General form of quadratic equation is  </u></em>

{ax^{2}+\mathrm{b} x+\mathrm{c}=0

And quadratic formula for getting roots of quadratic equation is

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

In our case b = 20, a = 1 and c = -15000

Calculating roots of the equation we get

\begin{array}{l}{x=\frac{-(20) \pm \sqrt{(20)^{2}-4(1)(-15000)}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{400+60000}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{60400}}{2}} \\\\ {x=\frac{-(20) \pm 245.764}{2 \times 1}}\end{array}

\begin{array}{l}{=>x=\frac{-(20)+245.764}{2 \times 1} \text { or } x=\frac{-(20)-245.764}{2 \times 1}} \\\\ {=>x=\frac{225.764}{2} \text { or } x=\frac{-265.764}{2}} \\\\ {=>x=112.882 \text { or } x=-132.882}\end{array}

As variable x represents width of the rectangular parking lot, it cannot be negative.

=> Width of the rectangular parking lot "x" = 112.882 feet  

=> Length of the rectangular parking lot = x + 20 = 112.882 + 20 = 132.882

Hence can conclude that dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet.

3 0
3 years ago
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