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cestrela7 [59]
3 years ago
6

find all the solutions of sec theta - 2 =0. It says use "or" as necessary and gives an example of kpi.

Mathematics
1 answer:
swat323 years ago
5 0
Sec t (theta) - 2 = 0
sec t = 2
1/ cos t = 2
cos t =  1/2
Answer:
t = π / 3 + 2 k π ,  or :  t = 5 π / 3 + 2 kπ,
k ∈ Z
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Restrict the domain of the quadratic function and find its inverse. Confirm the inverse relationship using composition. f(x) = 0
ANTONII [103]

The given function is

f(x)= 0.2 x²

Since f(x) will be defined for all real values of x.

So, Domain of f(x) will be ( x| x is a real number.)→This is set builder notation.

Finding the inverse of f(x):

y = 0.2 x²

→ x²= 5 y

→x = \pm\sqrt{5 y}→ → Inverse of f(x)

Replacing x by y and y by x,we get inverse of the given function

y = \pm\sqrt{5 x}→ →Domain x ≥ 0, x∈[0,∞]

Graph of function and its inverse are shown below.

8 0
3 years ago
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Find the 27th term in the arithmetic sequence of 59,56,53,50
statuscvo [17]

solution:

A_{n}=59-3(n-1)\\where, n=1,2,3,.....\\A1=59\\A2=56\\A3=53\\A_{n}=a+(n-1)dA27=59+(27-1)3\\=59-78\\=-19\\A27 term is -19

5 0
3 years ago
Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on
Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

  • If he is given with two kings.
  • If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by =  \frac{^{3}C_2 }{^{47}C_2}   {∵ one king is already there}

              =  \frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}           {∵ ^{n}C_r = \frac{n!}{r! \times (n-r)!} }

              =  \frac{3}{1081}  =  0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by =  \frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}

                                                                   =  \frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}

                                                                   =  \frac{6}{1081}  =  0.0055

Now, probability that he ends up with a full house = \frac{3}{1081} + \frac{6}{1081}

                                                                                    =  \frac{9}{1081} = <u>0.0083</u>.

3 0
3 years ago
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Higher Order Thinking The band
dolphi86 [110]
Do you know how to start
6 0
2 years ago
4n - 2n = 4<br><br>what's the answer for this ?
Mademuasel [1]
Hey there!
First, we will do the subtraction with the two like terms on the left. I know they're like terms because although they have different coefficients, they have the same variable.
We have:
2n=4
Divide both sides by 3 to get:
n = 2
Hope this helps!
5 0
3 years ago
Read 2 more answers
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