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Lorico [155]
3 years ago
14

If f(x)=3x^2 and g(x)=4x^3+1 , what is the degree of (f o g)(x)

Mathematics
1 answer:
pishuonlain [190]3 years ago
6 0

Answer D

...........................................

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Find all the complex square roots of W= 100 (cos 60 + I sin 60). Write the roots in polar form with theta in degrees.
serg [7]

Answer:

2√3+2i

hope it helps

please mark Brainliest

5 0
3 years ago
Help! Find m^KNL.<br> A. 264<br> B. 196<br> C. 184<br> D. 247
Misha Larkins [42]

Answer:

  A.  264

Step-by-step explanation:

First, we have to find the value of x. Then we can use that to find the required arc measure.

∠M = (1/2)(arc KN - arc LN)

60 = (1/2)((18x -6) -(5x +17)) = (1/2)(13x -23) . . . . substitute and simplify

120 = 13x -23 . . . . . . . multiply by 2

143 = 13x . . . . . . . . . . add 23

11 = x . . . . . . . . . . . . . . divide by 13

__

arc KNL = (arc KN) + (arc NL) = (18x -6) +(5x +17) = 23x +11

  = 23·11 +11

arc KNL = 264 . . . . degrees

5 0
3 years ago
Complete the steps for solving 7 = –2x2 + 10x.
Nikitich [7]

Answer:

1. -2

2. add 25/4

3. sub 25/2

4. 5/2

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A large electronic office product contains 2000 electronic components. Assume that the probability that each component operates
KIM [24]

Answer:

The probability is 0.971032

Step-by-step explanation:

The variable that says the number of components that fail during the useful life of the product follows a binomial distribution.

The Binomial distribution apply when we have n identical and independent events with a probability p of success and a probability 1-p of not success. Then, the probability that x of the n events are success is given by:

P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}

In this case, we have 2000 electronics components with a probability 0.005 of fail during the useful life of the product and a probability 0.995 that each component operates without failure during the useful life of the product. Then, the probability that x components of the 2000 fail is:

P(x)=\frac{2000!}{x!(2000-x)!}*0.005^{x}*(0.995)^{2000-x}     (eq. 1)

So, the probability that 5 or more of the original 2000 components fail during the useful life of the product is:

P(x ≥ 5) = P(5) + P(6) + ... + P(1999) + P(2000)

We can also calculated that as:

P(x ≥ 5) = 1 - P(x ≤ 4)

Where P(x ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4)

Then, if we calculate every probability using eq. 1, we get:

P(x ≤ 4) = 0.000044 + 0.000445 + 0.002235 + 0.007479 + 0.018765

P(x ≤ 4) = 0.028968

Finally, P(x ≥ 5) is:

P(x ≥ 5) = 1 - 0.028968

P(x ≥ 5) = 0.971032

3 0
3 years ago
I don't understand it at alll someone please give me the answer
goldenfox [79]
I hope this helps you

7 0
3 years ago
Read 2 more answers
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