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alexandr1967 [171]
3 years ago
14

Prove that root3 (cosec (20))-sec (20)=4

Mathematics
2 answers:
Tems11 [23]3 years ago
7 0
Root 3 /sin 20 - 1/cos 20

= (root3 cos 20 - sin 20 )/ (sin 20 cos 20)
(multiplying and dividing by 2)
=  2* (root3/2 cos 20 -1/2 sin 20)/ (sin 20 cos 20)

now sin 20 cos 20  = (2 sin 20 cos 20)/2 = (sin 40)/2
and writing root3/2 = sin 60, 1/2 = cos 60

= 2* (sin 60 cos 20 - cos 60 sin 20)/ (sin 40)/2

= 4* (sin (60-20))/sin 40

= 4* (sin 40)/sin 40
=4

Ask if any doubt in any step :)
Mashcka [7]3 years ago
3 0
Hello,

1) Rappels:

all angles are in degree.

cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
2sin(a)cos(a)=sin(2a)
sin(a)=cos(90-a)
cos(30)=√3/2
sin(30)=1/2
2)
√3 cos(20)-sin(20)=2(√3/2 cos(20)-1/2 sin(20)
=2*(cos(30) *sin(20)-sin(30) sin(20))=2 cos(30+20)=2 cos(50)

sin(20) cos(20)=1/2 cos(40)=1/2 cos(50)

<span>√3 (cosec (20))-sec (20) = √3 /sin(20)-1/cos(20)
=[√3*cos(20)-sin(20)]/(sin(20)cos(20))
=2 cos(50)/ (1/2cos(50))=4

</span>

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Any 10th grader solve it <br>for 50 points​
kkurt [141]

Answer:

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0  is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., a_{1}=A and common difference=D

The nth term can be written as

a_{n}=A+(n-1)D

pth term of given arithmetic progression is a

a_{p}=A+(p-1)D=a

qth term of given arithmetic progression is b

a_{q}=A+(q-1)D=b and

rth term of given arithmetic progression is c

a_{r}=A+(r-1)D=c

We have to prove that

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0

Now to prove LHS=RHS

Now take LHS

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)

=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)

=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)

=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}

=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}

=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

\neq 0

ie., RHS\neq 0

Therefore LHS\neq RHS

ie.,\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0  

Hence proved

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3 years ago
The manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .
Misha Larkins [42]

Answer:

0.1507 or 15.07%.

Step-by-step explanation:

We have been given that the manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .016 millimeters. To be acceptable the diameter needs to be between 21.97 and 22.03 millimeters.

First of all, we will find z-scores for data points using z-score formula.

z=\frac{x-\mu}{\sigma}, where,

z = z-score,

x = Sample score,

\mu = Mean,

\sigma = Standard deviation.

z=\frac{21.97-22}{0.016}

z=\frac{-0.03}{0.016}

z=-0.1875

Let us find z-score of data point 22.03.

z=\frac{22.03-22}{0.016}

z=\frac{0.03}{0.016}

z=0.1875

Using probability formula P(a, we will get:

P(-0.1875

P(-0.1875  

P(-0.1875

Therefore, the probability that a randomly selected ball bearing will be acceptable is 0.1507 or 15.07%.

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3 years ago
Need help right way asap
kari74 [83]

Answer:

A is 2 out of 6

Step-by-step explanation:

Probability of him pulling the one with parallel sides is 2/6 because there are only two out of 6 with parallel lines

7 0
1 year ago
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Expression for "the sum of a and 4'
Masteriza [31]
Answer: a + 4
explanation: sum means what comes from two numbers being added. so you just need to add a and 4
5 0
2 years ago
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Eva has £6.05 in her moneybox.
PIT_PIT [208]

Answer:

11 of 20p, 22 of 10p and 33 of 5p

Step-by-step explanation:

Eva has 20p, 10p and 5p coins, total of £6.05 = 605p

Let 20p=x, 10p=y, 5p=z

<u>Then</u>:

  • 20x + 10y + 5z = 605
  • y : x = 2 : 1 ⇒ x= y/2
  • y : z = 2 : 3 ⇒ z= 3y/2

<u>Rewriting the first equation considering next two:</u>

  • 10y + 10y + 7.5y = 605
  • 27.5y = 605
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<u>Answer:</u> 11 of 20p coins, 22 of 10p coins and 33 of 5p coins

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2 years ago
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