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grin007 [14]
3 years ago
5

Graph f(x)=2^x+1 and g(x)=−x+4 on the same coordinate plane.

Mathematics
2 answers:
12345 [234]3 years ago
8 0

Given f(x)= 2^x +1

g(x)= -x +4

We have to find the value of x which satisfy the condetion : f(x)= g(x)

Solution : let us place expression given for f(x) and g(x) equal to each other

This gives us : 2^x + 1 = -x+ 4

Let us bring all x terms on left side only

we add x on both sides this makes : 2^x +1 +x = -x+ 4 +x

on the right side -x and +x becomes a "0"

we get : 2^x + 1 + x = 4

* let us nopw bring all the numeric terms on left side only

subtract 1 from both sides : 2^x + 1 + x - 1 = 4- 1

on the left side +1 and -1 becomes a "0"

on the rigth side 4-1 becomes a "3"

equation look like : 2^x + x = 3

We can see the right side is an odd number

on the left side there is sum of two terms one of them is 2^x which is an even number always .

we know that sum of an even number with an even number result out an even number

but we want a result as an odd number (3)

this suggest that x should be an odd number .

Also the sum of 2^x and x is 3 so 2^x should be smaller than 3

the smallest value ( x being an integer )

2^x can have is : 2^1= 2 which is for x=1

let us check plugging x= 1 . 2^1 + 1 = 3 which is TRUE

Hence answer is x= 1

NeX [460]3 years ago
5 0

For graphing the functions, first we need to make table for finding ordered pairs.

f(x)= 2ˣ + 1 and g(x)= -x +4

After getting the points in each table, we need to plot them on graph paper and join the points for getting each graph.

Now, for finding the solution, we will look at the point where two graphs intersect with each other and here the point is (1,3)

So, x= 1

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perimeter = 20.9 units

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perimeter

perimeter = distance around two dimensional shape

= addition of all sides lengths

<h2>perimeter of the figure</h2><h2>= AB+BC+CD+AD</h2>

distance formula:

d =  \sqrt{(  x_{2} -  x_{1})  {}^{2}   + ( y_{2} -  y_{1}) {}^{2}  }

<h3>1) distance of AB</h3>

A(-3,0) B(2,4)

x1 = -3 x2 = 2

y1 = 0 y2 = 4

(substitute the values into the distance formula)

ab =  \sqrt{(2 - ( - 3)) {}^{2}  + (4 - 0) {}^{2} }

ab =  \sqrt{5 {}^{2} + 4 {}^{2}  }

ab =  \sqrt{41}AB = 6.4 units

<h3>2) distance of BC</h3>

B(2,4) C(3,1)

x1 = 2 x2 = 3

y1 = 4 y2 = 1

bc =  \sqrt{(3 - 2) {}^{2}  + (1 - 4) {}^{2} }

bc = \sqrt{1 {}^{2} + ( - 3) {}^{2}  }

bc =  \sqrt{10}

BC = 3.2 units

<h3>3) distance of CD</h3>

C(3,1) D(-4,-3)

x1 = 3 x2 = -4

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cd =   \sqrt{65}

CD = 8.1 units

<h3>4) distance of AD</h3>

A(-3,0) D(-4,-3)

x1 = -3 x2 = -4

y1 = 0 y2 = -3

ad =  \sqrt{(  - 4 - ( - 3)) {}^{2}  +  ( - 3 - 0) {}^{2} }

ad =  \sqrt{( - 1) {}^{2} + ( - 3) {}^{2}  }

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AD = 3.2 units

<h2>perimeter of figure</h2>

= AB+BC+CD+AD

= 6.4 + 3.2 + 8.1 + 3.2

= 20.9 units

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