If the domain of f(x) = x^2 + 1 is limuted to {0,1,2,3} what is the maximum value of the range?

Mathematics

1 answer:

katrin2010 [14]3 years ago

8 0

Answer: 10

Plug in the given x values and see which result is the largest

Plug in x = 0 ---> x^2+1 = 0^2 + 1 = 1

Plug in x = 1 ----> x^2 + 1 = 1^2 + 1 = 2

Plug in x = 2 ----> x^2 + 1 = 2^2 + 1 = 5

Plug in x = 3 ---> x^2 + 1 = 3^2 + 1 = 10

The range is therefore {1, 2, 5, 10} which is just the set of possible outputs.

We see that y = f(x) = 10 is the largest output of the range if the domain is restricted to the set {0, 1, 2, 3}

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Help pls 2,4 +0,8=........+1,21=........+1,78=........-5,14=

Mrrafil [7]

Lets write this out:-

2.4 + 0.8 = ________ + 1.21 = ______ + 1.78 = ______ - 5.14 = _____

So to solve d blanks we will do d following:-

2.4 + 0.8 = 3.2

Now lets write this out AGAIN.

2.4 + 0.8 = 3.2 + 1.21 = ______ + 1.78 = ______ - 5.14 = ____

Now lets solve again:-

3.2 + 1.21 = 4.41

Now lets write this out AGAIN.

2.4 + 0.8 = 3.2 + 1.21 = 4.41 + 1.78 = ______ - 5.14 = ____

Now lets solve again:-

4.41 + 1.78 = 6.19

Now lets write this out AGAIN.

2.4 + 0.8 = 3.2 + 1.21 = 4.41 + 1.78 = 6.19 - 5.14 = ____

Now lets solve again:-

6.19 - 5.14 = 1.05

Now lets write this out AGAIN.

2.4 + 0.8 = 3.2 + 1.21 = 4.41 + 1.78 = 6.19 - 5.14 = 1.05

So, 2.4 + 0.8 = 3.2 + 1.21 = 4.41 + 1.78 = 6.19 - 5.14 = 1.05

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7 0

3 years ago

What is the area of this figure?

Elanso [62]

First of all we can draw a parallel line to divide the figure into a triangle and a rectangle as shown in the figure. To find the area of our rectangle, remember that the area of a rectangle is length times width, so $A_{r} =lw$ . Since we know for our figure that the length and width of our rectangle are 13cm and 6cm respectively, lets replace those values in our formula to get its area:
$A _{r} =(13cm)(6cm)$
$A=78cm^{2}$
Similarly, the area of a triangle is one half times base times height, so $At=( \frac{1}{2})bh$ . Since we know that our base is 8cm and our height 6cm, lets replace those values in our equation to find the area of our triangle:
$A_{t} =( \frac{1}{2})(8cm)(6cm)$
$A_{t} =24cm^{2}$

Now the only thing left is add our areas:
$A_{total} =78cm^{2}+24cm^{2} =102cm^{2}$

We can conclude that the correct answer is <span>A. 102</span>