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Arisa [49]
3 years ago
6

Could someone please help me out? I've tried to solve this but I'm really having some trouble with it. Please try to explain it

in the easiest way that you can. (:
Write an expression for the area of the shaded region in its simplest form. Show all of your steps.

Mathematics
1 answer:
guajiro [1.7K]3 years ago
4 0

To find the area of the shaded region you need find the area of the shaded region and subtract the area of the unshaded region.

Area of a rectangle = width x length

A = (x + 10) x (2x + 5)

Next apply FOIL or First Outer Inner Last

A = (x * 2x) (x * 5) (10 * 2x) (10 * 5)

A= 2x2 + 5x + 20x + 50

A= 2x2 +25x +50

 

Area of a square=  sides2

A= (x + 1)2

A= (x+1) (x+1)

Next apply FOIL or First Outer Inner Last

A = (x *x) (1*x) (1*x) (1*1)

A = x2 + 1x + 1x +1

A= x2 + 2x +1

 

A= 2x2 +25x +50 - 2x2 +25x +50

A= 50x + 100





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Ms. Alma where is 86 kg she wants to reduce her body fat by 5% over the next month How many kilograms rounded to the nearest hun
djverab [1.8K]

Answer: 430 rounded so ur answer is 400

Step-by-step explanation:

you do 86 times 5 and you get 430 the you round it to the nearest hundred and you get 400 as ur answer

4 0
2 years ago
Read 2 more answers
Which expression can be used to model the phrase "the sum of three and a number"?
NeX [460]
The sum of three and a number is 3+x
3 0
3 years ago
Given that the product of 22 integers is equal to 1, prove that their sum is not equal to 0.
Kay [80]

1. If the product of these integers is to be 1, then all of them must be either 1 or -1.

2. Since the product is positive (+1), it must be that there are an *even* number of negative ones (-1), if any.

3. If the sum were 0 it would mean that the number of +1's must equal the number of -1's. So that means there would have to be exactly 22/2=11 of each.

4. But if there were 11 of each, that means the number of -1's would be *odd* and there's no way the product could be +1 (as stated in 2 above).

Hence, the sum is never 0, if the product of 22 integers is equal +1.

4 0
3 years ago
Read 2 more answers
PLEASE HELP ME ON THIS ONE I REALLY DONT KNOW HOW TO DO THIS AND I JUST CAME TO AMERICA SO I DONT KNOW HOW TO DO THESE THINGS CO
eimsori [14]

a: Answer is cuboid

b: top & base will be 5*6 = 30 in ^2

both right & left side wil be 5*2 = 10 in^2

front & back will be 6*2 = 12 in^2

Hope that will help you :)

5 0
3 years ago
Factories 54m3n + 81m4n2 b) 15x2y3z + 25x3y2z + 35x2y2z​
Oliga [24]

<u>Part a)</u>

Given the expression

54m^3n\:+\:81m^4n^2

Apply exponent rule:    a^{b+c}=a^ba^c

54m^3n\:+\:81m^4n^2=54m^3n+81m^3mnn     ∵ m^4n^2=m^3mnn

Rewrite 81 as 3 · 27

Rewrite 54 as 2 · 27

                             =2\cdot \:27m^3n+3\cdot \:27m^3mnn

Factor out the common term:   27m³n

                              =27m^3n\left(2+3mn\right)      

Therefore,

54m^3n\:+\:81m^4n^2=54m^3n+81m^3mnn=27m^3n\left(2+3mn\right)

<u>Part B)</u>

Given the expression

15x^2y^3z+25x^3y^2z+35x^2y^2z

Apply exponent rule:    a^{b+c}=a^ba^c

15x^2y^3z+25x^3y^2z+35x^2y^2z=15x^2y^2yz+25x^2xy^2z+35x^2y^2z

Rewrite as

                                               =3\cdot \:5y^2x^2zy+5\cdot \:5y^2x^2zx+7\cdot \:5y^2x^2z

Factor out common term 5y²x²z

                                                =5y^2x^2z\left(3y+5x+7\right)

Therefore,

15x^2y^3z+25x^3y^2z+35x^2y^2z=5y^2x^2z\left(3y+5x+7\right)

8 0
3 years ago
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