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madam [21]
3 years ago
11

Help please (inequalities)

Mathematics
2 answers:
Liula [17]3 years ago
7 0
The first one is the correct answer hope this helps have a nice day let me know if it works
zaharov [31]3 years ago
4 0

Answer:

The First one is the answer

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Which equation shows the product 2.4 x 10?<br><br><br><br><br><br><br>PLZ HELP :(​
Elina [12.6K]

The Answer:

2.4x10=24

there is one 0 in 10, so you move the decimal over by 1, resulting in 2.4 turning into 24

8 0
3 years ago
Read 2 more answers
Error analysis help me out
IrinaK [193]

Answer:

1 - sin² θ = 1 - ( 1 - cos² θ) = cos² θ

tan x csc x = (sin x / cos x) * 1 / sin x = cos x

Step-by-step explanation:

1 - sin² θ = 1 - ( 1 - cos² θ) = cos² θ

tan x csc x = (sin x / cos x) * 1 / sin x = cos x

5 0
4 years ago
Stefano accidentally dropped his sunglasses off the edge of a canyon as he was looking down. The height, h(t), in meters (as it
vlada-n [284]

Answer:

176 meters

Step-by-step explanation:

Stefano is at +10 elevation, the hiker is at -166.4

10 + 166.4 = 176.4 ≈ 176 meters

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3 years ago
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Select all of the following true statements if R = real numbers, N = natural numbers, and S = {10, 11, 12,...}.
Fofino [41]

The true statements are:

B. N⊂R

C. S⊂N

D. ∅⊂N

F. {10.11.12,...}⊆S

<h3>Subsets and Sets</h3>

Note that:

  • All real numbers are numbers that can be found on the real number line
  • Natural numbers are positive whole numbers
  • All natural numbers are subsets of real numbers
  • The null set (∅) is a subset of every set
  • A set is a proper subset of itself

Following the points highlighted above, the true statements in the given options re:

B. N⊂R

C. S⊂N

D. ∅⊂N

F. {10.11.12,...}⊆S

Learn more on subsets and sets here: brainly.com/question/13265691

#SPJ1

6 0
2 years ago
A​ quality-control inspector examined 100 light bulbs and found 7 defective. At this​ rate, how many defective bulbs would there
Vanyuwa [196]

we are given that out of 100 light bulbs, 7 of them were defective. We can assume that out of every 100 lightbulbs, 7 of them will be defective.

we can write this as a ratio of defective lightbulbs to total

7:100

we want to see how many defective there are when we have 3000 bulbs, to do this we can multiply both numbers by 30

7 * 30 = 210

100 * 30 = 3,000

our new ratio is now

210:3,000

so we are now able to say that in a lot of 3,000 lightbulbs, there would be 210 defective bulbs.

hope this helped!

5 0
3 years ago
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