I need a lot of help! Is 54.03 + 4.04 greater than 54 + 4.4?

Mathematics

1 answer:

Aliun [14]2 years ago

7 0

Answer:

54+4.4 is greater

Step-by-step explanation:

if u need explanation than please ask

You might be interested in

FIRST GET BRAINIEST PERIODTT TYSMMSMSMMSM

ValentinkaMS [17]

4+4+6+6=12+8=20

hope this helps

7 0

2 years ago

Read 2 more answers

If the original square had a side length of

irina [24]

Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is $x^{2} +11x+28$

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

$Length=(x+4)\ in$

$width=(x+7)\ in$

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

$A1=(x)(x)=x^{2}\ in^2$