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zalisa [80]
3 years ago
8

4)(x - 2) = 0" alt="(x - 3)(x + 4)(x - 2) = 0" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
AlexFokin [52]3 years ago
8 0

Answer:

Step-by-step explanation:

x-3 = 0

x = 3

x + 4 = 0

x = -4

x - 2 = 0

x = 2

x = 3, -4, 2

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2.8-5.3n - 4n + 5 =<br>plss help again <br>Thank you!<br>​
ra1l [238]

Answer:

7.8-9.3n

Step-by-step explanation:

2.8-5.3n - 4n + 5

Combine like terms

2.8 +5 -5.3n -4n

7.8-9.3n

4 0
2 years ago
Read 2 more answers
SH<br>What is the greatest common factor (GCF) of 35 and 28?<br>​
Brilliant_brown [7]

Answer: 7

Step-by-step explanation:

We found the factors and prime factorization of 28 and 35. The biggest common factor number is the GCF number. So the greatest common factor 28 and 35 is 7.

Hope this helps :)

6 0
3 years ago
Read 2 more answers
Please help!!! I have a chance to get $50 if I get all of these questions right!!
AnnyKZ [126]

Answer:

number 1 is c      number 2 is b,c,e  number 3 is b

Step-by-step explanation:

1: if you see something to the power you will always think of repeated multiplication off the bat. so you just multiply the coefficient by the amount of the exponent.

2: 1*1*1*1*1*1*1*1*1*1*1*1*1*1*1*1=1 no

2*2*2*2=16 yes

4*4=16 yes

8*8=64 no

16=16 yes

3: 3/4*3/4*3/4=27/64

hope this helps! have a nice day!

3 0
3 years ago
Hi!! Please help me with this question. I got an answer of 63.25 but it was incorrect and need to redo it. Please also explain h
matrenka [14]

Answer:

x ≈ 63.27

Step-by-step explanation:

Using the tangent ratio in the right triangle

tan49° = \frac{opposite}{adjacent} = \frac{EF}{DE} = \frac{x}{55} ( multiply both sides by 55 )

55 × tan49° = x , then

x ≈ 63.27 ( to the nearest hundredth )

4 0
2 years ago
We are given AB ∥ DE. Because the lines are parallel and segment CB crosses both lines, we can consider segment CB a transversal
ohaa [14]

Answer: AAA similarity.


Step-by-step explanation:  CB is the transversal for the parallel lines AB and DE, and so by transverse property, we have ∠CED ≅ ∠CBA. Similarly, CA acts as a tranversal for the same pair of parallel lines AB and DE and using the same property, we can have ∠CDE ≅ ∠CAB. Now, in triangles CED and ABC, we have

∠CED ≅ ∠CBA,

∠CDE ≅ ∠CAB

and

∠DCE ≅ ∠ACB [same angle]

Hence, by AAA (angle-angle-angle) similarity,

△CED ~ △ABC.

Thus, the correct option is AAA similarity.


8 0
3 years ago
Read 2 more answers
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