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stich3 [128]
2 years ago
12

Hey! please help i’ll give brainliest

Mathematics
2 answers:
alex41 [277]2 years ago
5 0
Pretty sure it’s A hope it helps
erma4kov [3.2K]2 years ago
3 0

Answer:

I think A is the answer

hope this helps

have a good day :)

Step-by-step explanation:

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Which trigonometric identity can be used to prove the given statement?<br> tan0 cos0 = sin0
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The graph shows a proportional relationship between the number of kilometers traveled by a bicycle (y) and the number of minutes
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Step-by-step explanation:

6 0
2 years ago
HELP ME ASAP PLEASE!!
lilavasa [31]

Answer:

a=45 b=65

Step-by-step explanation:

You know that all the angle measures of a triangle add to 180. So the equation would be angle a + angle b + 70 = 180. Subsitute and simplify and you get 16x -2 + 70 = 180. Subtract 70 from 180 to get 16x -2=110 add 2 to get 16x =110. Divide each side by 16 to gte x=7. Subsitute x = 7 into the equations for angle a and angle b to get the answers above

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2 years ago
Complete the table for the given rule. Y= x+2/3
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Answer:

decimal: x0.6 or fraction: 2/3

Step-by-step explanation:

3 0
3 years ago
Rewrite each statement using the appropriate mathematical language 1. There exists a number b belonging to the set B 2. Even num
Shtirlitz [24]

Answer:

1. b ∈ B 2. ∀ a ∈ N; 2a ∈ Z 3. N ⊂ Z ⊂ Q ⊂ R 4. J ≤ J⁻¹ : J ∈ Z⁻

Step-by-step explanation:

1. Let b be the number and B be the set, so mathematically, it is written as

b ∈ B.

2. Let  a be an element of natural number N and 2a be an even number. Since 2a is in the set of integers Z, we write

∀ a ∈ N; 2a ∈ Z

3. Let N represent the set of natural numbers, Z represent the set of integers, Q represent the set of rational numbers, and R represent the set of rational numbers.

Since each set is a subset of the latter set, we write

N ⊂ Z ⊂ Q ⊂ R .

4. Let J be the negative integer which is an element if negative integers. Let the set of negative integers be represented by Z⁻. Since J is less than or equal to its inverse, we write

J ≤ J⁻¹ : J ∈ Z⁻

4 0
3 years ago
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