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3241004551 [841]
3 years ago
10

Each side of a square is (x-5) units. Which expression can be used to represent the area of the square?

Mathematics
1 answer:
Cloud [144]3 years ago
6 0

Answer:

Area = (x - 5)^2   and    Area = x^2 - 10x + 25

Step-by-step explanation:

Given

Length = (x - 5)\ units

Required

Determine the area of the square

Area = Length * Length

Area = (x - 5) * (x - 5)

This can be expressed as

Area = (x - 5)^2

Or:

Area = (x - 5) * (x - 5)

Open brackets

Area = x(x -5) -5(x - 5)

Area = x^2 - 5x - 5x + 25

Area = x^2 - 10x + 25

Hence, the expressions are:

Area = (x - 5)^2   and    Area = x^2 - 10x + 25

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Please show work<br> 26 lbs of onion cost $127.40. How much would 18 lbs cost\
Irina18 [472]
127.40/26 = $4.90 per pound
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It would cost $88.20
5 0
3 years ago
Solve y = x - 5 if the domain is -3
DedPeter [7]

Answer:

y = -8

Step-by-step explanation:

The domain is the x values so x=-3

y = x-5

Substituting x=-3

y = -3-5

y=-8

6 0
3 years ago
There were several sunny days this year before my birthday. Since my birthday, there have been 171717 more sunny days. My birthd
Lesechka [4]

Answer:

66 sunny days before birthday

Step-by-step explanation:

Total sunny days after birthday = 17

Total sunny days in year = 83

Birthday was not sunny

Total sunny days before birthday = 83 -17

                                                       = 66 sunny days

Read more on Brainly.in - https://brainly.in/question/6937259#readmore

4 0
4 years ago
PLEASE ANSWER QUICKLY TY!
kkurt [141]
4 because you multiplied the 5 by 2 to get to the 10, so you have to do that to the 2. 2x2=4
7 0
3 years ago
The answer has to be a geometric proof. Thank you!
Mariulka [41]

Given data:

The given triangle in which AD is on perpendicular bisector on BC.

In triangle ABD and ACD.

\begin{gathered} \angle ADB=\angle\text{ADC}=90^{\circ} \\ BD=CD(\text{given)} \\ AD=AD\text{ (common)} \\ \Delta ABD\cong\Delta ACD(\text{SAS)} \end{gathered}

Simmilary triangle BED and triangle CED.

\begin{gathered} \angle BDE=\angle CDE \\ BD=CD \\ ED=ED \\ \Delta BED\cong\Delta CED(SAS) \end{gathered}

The fisr expression can be written as,

\begin{gathered} \Delta ABD\cong\Delta ACD \\ \Delta\text{ABE}+\Delta BED\cong\Delta ACE+\Delta\text{CED} \end{gathered}

Substitute CED in place of BED.

\begin{gathered} \Delta ABE+\Delta CED\cong\Delta ACE+\Delta CED \\ \Delta ABE\cong\Delta ACE \end{gathered}

Thus, the triangle ABE is congruent to trriangle ACE.

7 0
1 year ago
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