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Dmitry_Shevchenko [17]
3 years ago
5

Alisa says it is easier to compare the numbers in set A 45,760-1,025,680 than set B492,111-409,867

Mathematics
1 answer:
Marysya12 [62]3 years ago
3 0
What is the question?
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HELP ME PLEASEEEEEEEEEEE
Alborosie

Answer:

2

Step-by-step explanation:

3x2=6

s=2

Plz brainly and like :D

5 0
3 years ago
Read 2 more answers
2〖sen〗^2 x+3 senx+1=0
KonstantinChe [14]

2 sin²(<em>x</em>) + 3 sin(<em>x</em>) + 1 = 0

(2 sin(<em>x</em>) + 1) (sin(<em>x</em>) + 1) = 0

2 sin(<em>x</em>) + 1 = 0   OR   sin(<em>x</em>) + 1 = 0

sin(<em>x</em>) = -1/2   OR   sin(<em>x</em>) = -1

The first equation gives two solution sets,

<em>x</em> = sin⁻¹(-1/2) + 2<em>nπ</em> = -<em>π</em>/6 + 2<em>nπ</em>

<em>x</em> = <em>π</em> - sin⁻¹(-1/2) + 2<em>nπ</em> = 5<em>π</em>/6 + 2<em>nπ</em>

(where <em>n</em> is any integer), while the second equation gives

<em>x</em> = sin⁻¹(-1) + 2<em>nπ</em> = -<em>π</em>/2 + 2<em>nπ</em>

2 cot(<em>x</em>) sec(<em>x</em>) + 2 sec(<em>x</em>) + cot(<em>x</em>) + 1 = 0

2 sec(<em>x</em>) (cot(<em>x</em>) + 1) + cot(<em>x</em>) + 1 = 0

(2 sec(<em>x</em>) + 1) (cot(<em>x</em>) + 1) = 0

2 sec(<em>x</em>) + 1 = 0   OR   cot(<em>x</em>) + 1 = 0

sec(<em>x</em>) = -1/2   OR   cot(<em>x</em>) = -1

cos(<em>x</em>) = -2   OR   tan(<em>x</em>) = -1

The first equation has no (real) solutions, since -1 ≤ cos(<em>x</em>) ≤ 1 for all (real) <em>x</em>. The second equation gives

<em>x</em> = tan⁻¹(-1) + <em>nπ</em> = -<em>π</em>/4 + <em>nπ</em>

<em />

sin(<em>x</em>) cos²(<em>x</em>) = sin(<em>x</em>)

sin(<em>x</em>) cos²(<em>x</em>) - sin(<em>x</em>) = 0

sin(<em>x</em>) (cos²(<em>x</em>) - 1) = 0

sin(<em>x</em>) (-sin²(<em>x</em>)) = 0

sin³(<em>x</em>) = 0

sin(<em>x</em>) = 0

<em>x</em> = sin⁻¹(0) + 2<em>nπ</em> = 2<em>nπ</em>

<em />

2 cos²(<em>x</em>) + 2 sin(<em>x</em>) - 12 = 0

2 (1 - sin²(<em>x</em>)) + 2 sin(<em>x</em>) - 12 = 0

-2 sin²(<em>x</em>) + 2 sin(<em>x</em>) - 10 = 0

sin²(<em>x</em>) - sin(<em>x</em>) + 5 = 0

Using the quadratic formula, we get

sin(<em>x</em>) = (1 ± √(1 - 20)) / 2 = (1 ± √(-19)) / 2

but the square root contains a negative number, which means there is no real solution.

2 csc²(<em>x</em>) + cot²(<em>x</em>) - 3 = 0

2 (cot²(<em>x</em>) + 1) + cot²(<em>x</em>) - 3 = 0

3 cot²(<em>x</em>) - 1 = 0

cot²(<em>x</em>) = 1/3

tan²(<em>x</em>) = 3

tan(<em>x</em>) = ± √3

<em>x</em> = tan⁻¹(√3) + <em>nπ</em>  OR   <em>x</em> = tan⁻¹(-√3) + <em>nπ</em>

<em>x</em> = <em>π</em>/3 + <em>nπ</em>   OR   <em>x</em> = -<em>π</em>/3 + <em>nπ</em>

7 0
3 years ago
Use mental math to find which has a greater product 5×50or 5×500
Dennis_Churaev [7]
I think the answer is 5×500
7 0
3 years ago
Read 2 more answers
How do you answer this?
lidiya [134]
(a + b)^3 = a^3 + 3a^2b + 3ab^2+ b^3 
(a +(- b))^3 = (a-b)^2 =  a^3 - 3a^2b + 3ab^2- b^3  
7 0
3 years ago
What is the value of the expression?
lora16 [44]
-square root 3 over 3
 hope this helps.
7 0
3 years ago
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