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SpyIntel [72]
4 years ago
5

18. What value of b will make the equation have no solution? 6x + 4 = 4bx – 2

Mathematics
1 answer:
MrRissso [65]4 years ago
5 0

Answer: B is what 6 x +4 = to 4bx - 2 is 6x plus 4 = to 24 then 4bx - 2 is  YA so b is Y

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2xsquared + 19x -6 = 180
Karolina [17]

Answer:

Step-by-step explanation:

7 0
4 years ago
Help please and ill mark you brainliest :)
rodikova [14]

Answer:

n=100+0.05(460)

Step-by-step explanation:

She earns 100 dollars all the time. so its 100+ something

she also earns 5% of the tips, which in that case were 460$

you can repersent that as 0.05(460)

so: 100+0.05(460)

4 0
3 years ago
A set of piano prices distributed at a mean of 3000 and a standard deviation of 200 dollars. An electric piano has a price of 25
lana [24]

Answer:

The proportion of piano prices higher than the electric piano is 98.3%

Step-by-step explanation:

The first thing to do here is to calculate the standard score of the price of the electric piano given.

Mathematically, this is

z-score = (x-mean)/SD

where mean is 3000 and SD is 200, x is 2576

z-score = (2576-3000)/200 = -2.12

Now we proceed to calculate the probability of this z-score

The probability we are trying to calculate is

P( x > $2576) or simply P ( z > -2.12)

Using standard score probability calculator or table, we have

P(x>2576) = 1 - P(x<2576)

But, P(x<2576) = 0.017003

P(x>2576) = 1 - P(x<2576) = 0.983

This is same as 98.3%

4 0
3 years ago
#12. How is the answer 48? How do I get it?
VashaNatasha [74]
You do 9+3 which equals 12 then you multiply that by 4
3 0
3 years ago
Read 2 more answers
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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