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aniked [119]
3 years ago
12

True or false. The equation csc^2x-1=cot^2x is an identity.

Mathematics
2 answers:
slava [35]3 years ago
5 0

Answer:

True

Step-by-step explanation:

Given: \csc^2x-1=\cot^2x

To find: False or True

Taking left side of equation

\Rightarrow \csc^2x-1

\Rightarrow \dfrac{1}{\sin^2}-1                          \because \csc x=\dfrac{1}{\sin x}

\Rightarrow \dfrac{1-\sin^2x}{\sin^2}

\Rightarrow \dfrac{\cos^2x}{\sin^2}                          \because 1-\sin^2x=\cos^2

\Rightarrow \cot^2x

Thus, The left hand side is equal to right ahnd side.

Hence, The statement is true.

KatRina [158]3 years ago
3 0
Csc² x - 1 = cot² x
1/ sin² x - 1 = cot² x
( 1 - sin² x ) / sin² x = cot² x
cos² x / sin² x = cot² x
cot² x = cot² x
Answer: True.
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During a race, a runner’s time was 4 seconds at the 36-meter mark and 6 seconds at the 57-meter mark. What was the runner’s aver
Arturiano [62]

Answer:

E

Step-by-step explanation:

57 - 36 =  21              6 - 4 = 2 seconds

21 ÷ 2 = 10.5

6 0
3 years ago
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Suppose you toss a fair coin 10 times, let X denote the number of heads. (a) What is the probability that X=5? (b) What is the p
zubka84 [21]

Answer:  The required answers are

(a) 0.25,    (b) 0.62,    (c) 6.

Step-by-step explanation:  Given that we toss a fair coin 10 times and X denote the number of heads.

We are to find

(a) the probability that X=5

(b) the probability that X greater or equal than 5

(c) the minimum value of a such that P(X ≤ a) > 0.8.

We know that the probability of getting r heads out of n tosses in a toss of coin is given by the formula of binomial distribution as follows :

P(X=r)=^nC_r\left(\dfrac{1}{2}\right)^r\left(\dfrac{1}{2}\right)^{n-r}.

(a) The probability of getting 5 heads is given by

P(X=5)\\\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}\\\\\\=\dfrac{10!}{5!(10-5)!}\dfrac{1}{2^{10}}\\\\\\=0.24609\\\\\sim0.25.

(b) The probability of getting 5 or more than 5 heads is

P(X\geq 5)\\\\=P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}+^{10}C_6\left(\dfrac{1}{2}\right)^6\left(\dfrac{1}{2}\right)^{10-6}+^{10}C_7\left(\dfrac{1}{2}\right)^7\left(\dfrac{1}{2}\right)^{10-7}+^{10}C_8\left(\dfrac{1}{2}\right)^8\left(\dfrac{1}{2}\right)^{10-8}+^{10}C_9\left(\dfrac{1}{2}\right)^9\left(\dfrac{1}{2}\right)^{10-9}+^{10}C_{10}\left(\dfrac{1}{2}\right)^{10}\left(\dfrac{1}{2}\right)^{10-10}\\\\\\=0.24609+0.20507+0.11718+0.04394+0.0097+0.00097\\\\=0.62295\\\\\sim 0.62.

(c) Proceeding as in parts (a) and (b), we see that

if a = 10, then

P(X\leq 0)=0.00097,\\\\P(X\leq 1)=0.01067,\\\\P(X\leq 2)=0.05461,\\\\P(X\leq 3)=0.17179,\\\\P(X\leq 4)=0.37686,\\\\P(X\leq 5)=0.62295,\\\\P(X\leq 6)=0.82802.

Therefore, the minimum value of a is 6.

Hence, all the questions are answered.

3 0
3 years ago
The following absolute value inequality is an example of a conjunction.
umka2103 [35]
The answers is false
7 0
3 years ago
Ray FL bisects ∠AFM. m∠LFM = (11x+4), m∠AFL = (12x - 2).
Ne4ueva [31]

Answer:

AFM = 140

LFM = 70

Step-by-step explanation:

Here, we are to calculate LFM and AFM

Since AFM was bisected, then LFM + AFL is AFM

and also AFL = LFM

Thus;

11x + 4 = 12x -2

12x-11x = 4 + 2

x = 6

AFM = AFL + LFM = 11x + 4 + 12x-2 = 23x + 2

Substitute x = 6

AFM = 23(6) + 2 = 140

LFM = 11x + 4 = 11(6) + 4 = 66 + 4 = 70

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