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daser333 [38]
3 years ago
6

What is a gift that is thoughtful and does not cost any money

Mathematics
2 answers:
Brums [2.3K]3 years ago
5 0
Breakfast in bed



Homemade card

Fudgin [204]3 years ago
4 0
I gift that is thoughtful is your love you don't need anything valuable to make the people you love happy.It doesn't cost money because God gave this power and gift to everyone :)
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Pi &  Phi in the human body. nautilus shells, Sunflowers, Seahorses, Pine Cones, Fibonacci Sequence etc.
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3 years ago
Need help with 6,7,8,9,10,11,12,15
Ainat [17]

Answer:

6. 1.01 x -10^5

7. 6 x -10^14

8. 550,000

9. 60,700,000

10. 0.000204

11. 0.0004

12. 7,000,000,000,000 > 3,500,000,000

     7,000,000,000,000 (or 7 x 10^12) is greater by 2,000 times

15. 10^3 and 10^4

7 0
3 years ago
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What are the solutions of the equation 2x2−x=32x2-x=3?
klio [65]
I think it'd be x=4 but I'm not 100% sure
8 0
3 years ago
A car traveled 11.5 miles in 15 minutes. How many miles per hour was it traveling?
polet [3.4K]

Answer:

46 miles per hour

Step-by-step explanation:

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So, if it travelled 11.5 miles in 15 minutes, in an hour, it travelled 11.5 * 4 = 46 miles per hour.

8 0
3 years ago
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1. cot x sec4x = cot x + 2 tan x + tan3x
Mars2501 [29]
1. cot(x)sec⁴(x) = cot(x) + 2tan(x) + tan(3x)
    cot(x)sec⁴(x)            cot(x)sec⁴(x)
                   0 = cos⁴(x) + 2cos⁴(x)tan²(x) - cos⁴(x)tan⁴(x)
                   0 = cos⁴(x)[1] + cos⁴(x)[2tan²(x)] + cos⁴(x)[tan⁴(x)]
                   0 = cos⁴(x)[1 + 2tan²(x) + tan⁴(x)]
                   0 = cos⁴(x)[1 + tan²(x) + tan²(x) + tan⁴(4)]
                   0 = cos⁴(x)[1(1) + 1(tan²(x)) + tan²(x)(1) + tan²(x)(tan²(x)]
                   0 = cos⁴(x)[1(1 + tan²(x)) + tan²(x)(1 + tan²(x))]
                   0 = cos⁴(x)(1 + tan²(x))(1 + tan²(x))
                   0 = cos⁴(x)(1 + tan²(x))²
                   0 = cos⁴(x)        or         0 = (1 + tan²(x))²
                ⁴√0 = ⁴√cos⁴(x)      or      √0 = (√1 + tan²(x))²
                   0 = cos(x)         or         0 = 1 + tan²(x)
         cos⁻¹(0) = cos⁻¹(cos(x))    or   -1 = tan²(x)
                 90 = x           or            √-1 = √tan²(x)
                                                         i = tan(x)
                                                      (No Solution)

2. sin(x)[tan(x)cos(x) - cot(x)cos(x)] = 1 - 2cos²(x)
              sin(x)[sin(x) - cos(x)cot(x)] = 1 - cos²(x) - cos²(x)
   sin(x)[sin(x)] - sin(x)[cos(x)cot(x)] = sin²(x) - cos²(x)
                               sin²(x) - cos²(x) = sin²(x) - cos²(x)
                                         + cos²(x)              + cos²(x)
                                             sin²(x) = sin²(x)
                                           - sin²(x)  - sin²(x)
                                                     0 = 0

3. 1 + sec²(x)sin²(x) = sec²(x)
           sec²(x)             sec²(x)
      cos²(x) + sin²(x) = 1
                    cos²(x) = 1 - sin²(x)
                  √cos²(x) = √(1 - sin²(x))
                     cos(x) = √(1 - sin²(x))
               cos⁻¹(cos(x)) = cos⁻¹(√1 - sin²(x))
                                 x = 0

4. -tan²(x) + sec²(x) = 1
               -1               -1
      tan²(x) - sec²(x) = -1
                    tan²(x) = -1 + sec²
                  √tan²(x) = √(-1 + sec²(x))
                     tan(x) = √(-1 + sec²(x))
            tan⁻¹(tan(x)) = tan⁻¹(√(-1 + sec²(x))
                             x = 0
5 0
3 years ago
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