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Korolek [52]
3 years ago
6

PLZZZ help with this its pretty simple!

Mathematics
2 answers:
Zolol [24]3 years ago
7 0

Answer:

D

Step-by-step explanation:

hope this helps

My name is Ann [436]3 years ago
3 0

Answer:

D) 4x² - 9

Step-by-step explanation:

(2x + 3)(2x - 3)

4x² - 6x + 6x - 9

4x² - 9

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THis is due in 28 minutes help ASAP
Virty [35]

Answer:

I promise you that it is 120,000cm

Step-by-step explanation:

6 0
3 years ago
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Has anyone taken the math kangaroo competition
kirill [66]
No i haven’t i’m sorry,what even is that tho
6 0
3 years ago
Harlon recorded this set of data, which contains an outlier. 163, 97, 184, 199, 169, 175 What is the range of this set of data?
maks197457 [2]
Hey there!

To find the range, first put the numbers in order from smallest to largest. 
97, 163, 169, 175, 184, 199

Then take the largest number, 199, and subtract the smallest number (in this case, the outlier) which is 97.

199-97=102

Thanks for using Brainly! Hope this helps :)


7 0
3 years ago
Read 2 more answers
At the city museum, child admission is $5.20 and adult admission is $9.50. On Friday,133 tickets were sold for a total sales of
fredd [130]

Answer:

55 adult tickets were sold on Friday.

Step-by-step explanation:

A linear system will be used to model the system.

Let c be the number of child tickets and a be the number of adult tickets

Then according to given statements

a+c = 133\ \ \ \ Eqn\ 1\\9.50a+5.20c = 928.10\ \ \ Eqn\ 2

From equation 1

a = 133-c

Putting this in equation 2

9.50(133-c)+5.20c = 928.10\\1263.5-9.50c+5.20c=928.10\\1263.5-4.3c = 928.10\\-4.3c = 928.10-1263.5\\-4.3c=-335.4\\\frac{-4.3c}{-4.3} = \frac{-335.4}{-4.3}\\c = 78

Putting c = 78 in equation 1

a+78 = 133\\a = 133-78\\a = 55

Hence,

55 adult tickets were sold on Friday.

4 0
3 years ago
Simplify the following:<br> 1) √12 =
goldenfox [79]

Answer:

2\sqrt{3}

Step-by-step explanation:

Using the rule of radicals

\sqrt{a} × \sqrt{b} ⇔ \sqrt{ab} , then

\sqrt{12}

= \sqrt{4(3)}

= \sqrt{4} × \sqrt{3}

= 2\sqrt{3}

4 0
2 years ago
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