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STALIN [3.7K]
2 years ago
7

Stella buys 5 pineapples for $4.50. How much would she pay if she bought 7 pineapples? (hint: find the unit rate first then mult

iply)
*
Mathematics
2 answers:
Nezavi [6.7K]2 years ago
7 0
4.50 divided by 5 is .90 so .90 times 7 is 6.3
Sphinxa [80]2 years ago
5 0

Answer:

Step-by-step explanation:

4.50 divides by 5 than times the answer by 7

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-10n + 24 + 30n = -36<br> Solve this equation
vladimir1956 [14]

Step-by-step explanation:

- 10n + 24 + 30n =  - 36

Collect like terms and simplify

[tex] -10n+30n=-36-24 \\ 20n = -60 [/tex]

Divide both sides of the equation by 20

[tex] 20n/20 = -60/20 \\ n=-3 [/tex]

8 0
3 years ago
Read 2 more answers
Michelle buys a rug for $36 and sells it for $90. What was the value of her markup, expressed as a percent?
Rama09 [41]
Her markup was about 60%.
4 0
3 years ago
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F(x) = -x^3+ 6x – 7 at x = 2
vodomira [7]

Answer:

f(2)= -4

Step-by-step explanation:

f(x)= -x^3 + 6x - 7

f(2) = -2^3 + 6*2 - 7

f(2)= -8 +12 -7

f(2)= 3-7

f(2)= -4

4 0
2 years ago
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
Suppose that commercial airplane crashes in a certain country occur at the rate of 2.5 per year. what is the probability that fo
lesya [120]
The poisson distribution applies here.
\lambda=2.5
P(X=4)=\frac{e^{-2.5}\times2.5^{4}}{4!}=0.134
4 0
3 years ago
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