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3 years ago

Determine the measure of

Mathematics

2 answers:

vesna_86 [32]3 years ago

7 0

Answer:

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Rufina [12.5K]3 years ago

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Prove that: [1 + 1/tan²theta] [1 + 1/cot² thata] = 1/(sin²theta - sin⁴theta]

Stels [109]

Step-by-step explanation:

<h3><u>Given :-</u></h3>

[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]

<h3><u>Required To Prove :-</u></h3>

[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)

<h3><u>Proof :-</u></h3>

On taking LHS

[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]

We know that

Tan θ = 1/ Cot θ

and

Cot θ = 1/Tan θ

=> (1+Cot²θ)(1+Tan²θ)

=> (Cosec² θ) (Sec²θ)

Since Cosec²θ - Cot²θ = 1 and

Sec²θ - Tan²θ = 1

=> (1/Sin² θ)(1/Cos² θ)

Since , Cosec θ = 1/Sinθ

and Sec θ = 1/Cosθ

=> 1/(Sin²θ Cos²θ)

We know that Sin²θ+Cos²θ = 1

=> 1/[(Sin²θ)(1-Sin²θ)]

=> 1/(Sin²θ-Sin²θ Sin²θ)

=> 1/(Sin²θ - Sin⁴θ)

=> RHS

=> LHS = RHS

<u>Hence, Proved.</u>

<h3><u>Answer:-</u></h3>

[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)

<h3><u>Used formulae:-</u></h3>

→ Tan θ = 1/ Cot θ

→ Cot θ = 1/Tan θ

→ Cosec θ = 1/Sinθ

→ Sec θ = 1/Cosθ

<h3><u>Used Identities :-</u></h3>

→ Cosec²θ - Cot²θ = 1

→ Sec²θ - Tan²θ = 1

→ Sin²θ+Cos²θ = 1

Hope this helps!!

7 0

3 years ago

Estimate a 15% tip for a 26.80 meal

lina2011 [118]

26.80 x .15 gives you the answer, which is $4.02 in tip

7 0

3 years ago

Read 2 more answers

Please help ASAP on question b

max2010maxim [7]

Answer:

13.9cm

Step-by-step explanation:

13^2 = 169

5^2 = 25

169 + 25

= 194

13.9

8 0

3 years ago

Read 2 more answers

A detergent company periodically tests its products for variations in the fill weight. To do this, the company uses x-bar and R

maw [93]

Answer:

10.9361

Step-by-step explanation:

The lower control limit for xbar chart is

xdoublebar-A2(Rbar)

We are given that A2=0.308.

xdoublebar=sumxbar/k

Rbar=sumR/k

xbar R

5.8 0.42

6.1 0.38

16.02 0.08

15.95 0.15

16.12 0.42

6.18 0.23

5.87 0.36

16.2 0.4

Xdoublebar=(5.8+6.1+16.02+15.95+16.12+6.18+5.87+16.2)/8

Xdoublebar=88.24/8

Xdoublebar=11.03

Rbar=(0.42+0.38+0.08+0.15+0.42+0.23+0.36+0.4)/8

Rbar=2.44/8

Rbar=0.305

The lower control limit for the x-bar chart is

LCL=xdoublebar-A2(Rbar)

LCL=11.03-0.308*0.305

LCL=11.03-0.0939

LCL=10.9361

8 0

3 years ago

A cylinder shaped can needs to be constructed to hold 400 cubic centimeters of soup. The material for the sides of the can costs

LenKa [72]

Answer:

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

Step-by-step explanation:

Volume of the Cylinder=400 cm³

Volume of a Cylinder=πr²h

Therefore: πr²h=400

$h=\frac{400}{\pi r^2}$

Total Surface Area of a Cylinder=2πr²+2πrh

Cost of the materials for the Top and Bottom=0.06 cents per square centimeter

Cost of the materials for the sides=0.03 cents per square centimeter

Cost of the Cylinder=0.06(2πr²)+0.03(2πrh)

C=0.12πr²+0.06πrh

Recall: $h=\frac{400}{\pi r^2}$

Therefore:

$C(r)=0.12\pi r^2+0.06 \pi r(\frac{400}{\pi r^2})$

$C(r)=0.12\pi r^2+\frac{24}{r}$

$C(r)=\frac{0.12\pi r^3+24}{r}$

The minimum cost occurs when the derivative of the Cost =0.

$C^{'}(r)=\frac{6\pi r^3-600}{25r^2}$

$6\pi r^3-600=0$

$6\pi r^3=600$

$\pi r^3=100$

$r^3=\frac{100}{\pi}$

$r^3=31.83$

r=3.17 cm

Recall that:

$h=\frac{400}{\pi r^2}$

$h=\frac{400}{\pi *3.17^2}$

h=12.67cm

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

3 0

3 years ago