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3 years ago

How do variables help you model real-world situations?

2 answers:

6 0

Variables help you model real-world situations by allowing you to assign certain symbols to a variable or factor. You can easily identify what are the dependent variables and independent variables. Also, you can easily know what to vary to know how much will change to an independent variable. You can predict what most likely happen in the real world.

notka56 [123]3 years ago

3 0

A relationship between two quantities is often expressed as an equation in two variables. For instance, the equation:

$y=3-4x$

is an equation in two variables x and y. Each order pair (a, b) will be a solution of this equation if and only if makes the equation to be true. That is, the ordered pair (1, -1) is a solution because:

$-1=3-4(1) \ is \ true$

In this way, the graph of an equation can help you see relationships between real-life quantities. So, a real-world situation could be the life expectancy of a child, at birth, in the United States for selected years. So, if you are a researcher of a Research Institute you need to establish a relationship between the variable year and the variable life expectancy to know the life expectancy in each year.

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A square building with an area of 225 m2 has a garden surrounding it that has an equal width on all sides. The area of the garde

xenn [34]

Answer:

The width of the garden is 1.16 m

Step-by-step explanation:

step 1

Find the area of the garden

To find out the area of the garden multiply by 1/3 the area of the building

$225(\frac{1}{3})=75\ m^2$

step 2

Find the length side of the square building

The area of a square is

$A=b^2$

where

b is the length side of the square

we have

$A=225\ m^2$

$b^2=225\\b=15\ m$

step 3

Find the width of the garden

Let

x ----> the width of the garden

we know that

The area of the building plus the area of the garden is equal to

$(15+2x)^2=225+75$

solve for x

$225+60x+4x^2=225+75\\4x^2+60x-75=0$

Solve the quadratic equation by graphing

using a graphing tool

The solution is x=1.16 m

see the attached figure

therefore

The width of the garden is 1.16 m

Find the exact value

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$

is equal to

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

in this problem we have

$4x^2+60x-75=0$

$a=4\\b=60\\c=-75$

substitute in the formula

$x=\frac{-60\pm\sqrt{60^{2}-4(4)(-75)}} {2(4)}$

$x=\frac{-60\pm\sqrt{4,800}} {8}$

$x=\frac{-60\pm40\sqrt{3}} {8}$

$x=\frac{-60+40\sqrt{3}} {8}\ m$ ----> exact value

8 0

3 years ago

Simplify r^6 • r^5 type your answer in the box

TEA [102]

Answer:

r^11

Step-by-step explanation:

5 0

3 years ago

Please help solve Problem

Vinvika [58]

Hey! just so you know you put it upside down but the answer is 23.

3 0

3 years ago

Read 2 more answers

How do you use the associative property to write an expression equivalent to (w+9)+3

m_a_m_a [10]

The equivalent expression for the given expression (w+9)+3 using associative property is w + 12

Solution:

Given that, expression is ( w + 9 ) + 3

We have to find how do we can use the associative property to write an expression equivalent to ( w + 9 ) + 3

The associative property states that you can add or multiply regardless of how the numbers are grouped

By above definition, we get

(a + b) + c = a + (b + c)

So, now let us apply the above property to the given expression,

Then, ( w + 9 ) + 3 = w + ( 9 + 3 ) = w + 12

Hence, the equivalent expression for the given expression is w + 12

5 0

3 years ago

Consider the equation 2/3x-9-2x+2=1.

bogdanovich [222]

The one is the same everywhere so we ignore it.

$\frac 2 {3} x -9 -2x+2 = (\frac 2 3 - \frac 6 3) x - 9+2= -\frac 4 3 x -7$

Choice B

6 0

3 years ago

Read 2 more answers