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Naddika [18.5K]
3 years ago
15

Expand (2x+5)give answer plz​

Mathematics
1 answer:
emmasim [6.3K]3 years ago
7 0

Answer:

2(x+5)

not sure but this is what I got

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A square board was attached to a wall. The board was divided by four squares and each square was colored either black or white.
PtichkaEL [24]

Answer:

24 combinations

Step-by-step explanation:

For this problem, you use factorials. Factorials are when you multiply every number below it except zero. So like the factorial of 3 or 3! is 3*2*1 is 6.

Factorials are used to figure out how many combinations there are of something. So for you, we have 4 different quadrants so we write that as 4! (! is the sign for factorials.)

By the way, welcome to brainly.  

4 0
3 years ago
I need help with an Algebra question. The question is in the picture. Thank you <3
olchik [2.2K]
The domain is the set of allowed inputs, in this case t values. The smallest t value allowed is t = 0. The largest is t = 165. So that's why the domain is 0 \le t \le 165

-------------------------------

The range is 0 \le H \le 40000 since H = 0 is the smallest output of the function and H = 40,000 is the largest output. Like the domain, the range is the set of possible outputs of a function.

4 0
3 years ago
Which of the following is the correct notation for the complex number √-64-12
Liono4ka [1.6K]
The answer would be -12+8i
6 0
3 years ago
Please help me !!!
Pie
3. 5mn ( 2m + 6n + 1 )
4. 4a^3b^2c ( 18a^2b - 8c )
5. (X-15)(X+15)
6. -4(y-4.5)(y+4.5)
8. 2X(X-10)(X+10)
i am not sure if no. 8 is right
8 0
3 years ago
Read 2 more answers
Find the sum of the first 40 terms of a geometric sequence where the first term is 16 and the common ratio is 1.1. 704 7,081.480
Lilit [14]

Step-by-step explanation:

The nth term of a geometric progression can be determined by using the formula:

Tn=arn−1

where: a = first term and r = common ratio

Substitute the given values of first term and common ratio into the formula:

Tn=arn−1

T5=(40)(0.5)5−1

T5=(40)(0.5)4

T5=(40)(0.0625)

T5=2.5

8 0
3 years ago
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