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Ebi, Jose, Derell, and Asami measured their heights. Ebi's height was 2.5 cm greater than Jose's height. Jose's height was 3.1 c

irga5000 [103]

Start with assigning each person with a variable to represent their height

Ebi: e
Jose: j
Derell: d
Asami: a

Ebi'd height was 2.5 cm greater than Jose's height

j + 2.5 = e

Jose's height was 3.1 cm greater than Derell's

d + 3.1 = j

Derell's height is 0.4 cm less than Asami's height

a - 0.4 = d

Ebi is 162.5 cm tall

e = 162.5

So, plug in 162.5 into any of the above equations were there is a variable of e

j + 2.5 = e

j + 2.5 = 162.5

Subtract 2.5 from both sides of the equation

j = 160 cm

Jose's height is 160 cm

Now, plug in 160 into any of the above equations where there is a j

d + 3.1 = j

d + 3.1 = 160

Subtract 3.1 from both sides of the equation

d = 156.9 cm

Derell's height 156.9 cm

so, plug in 156.9 into any of the above equations where there is a d

a - 0.4 = d

a - 0.4 = 156.9

Add 0.4 on both sides of the equation

a = 157.3 cm

Asami's height is 157.3 cm

7 0

3 years ago

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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>Solution:</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

The area of rectangle is given as:

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

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

General form of quadratic equation is

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

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

Find the length of segment EG. Show all the work

svp [43]

Answer:

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I don't know the answear sorry

5 0

2 years ago

I need a models or visuals for that problem

Leya [2.2K]

Answer:I will be a mode

Step-by-step explanation:

4 0

3 years ago

What is 4,321,109,432 rounded to the nearest ten million

SOVA2 [1]

To round to the nearest ten million, you look at the millions digit. In this case, the millions digit is 1, which is less than 5, so you round the number down to 4,320,000,000.

5 0

3 years ago

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