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3 years ago

How does cot(x)sin(x)=cos^3(x)+cos(x)sin^2(x)?

Mathematics

1 answer:

Svetllana [295]3 years ago

5 0

$\cot(x)\sin(x)=\cos^3(x)+\cos(x)\sin^2(x)$

$R=\cos(x)[\cos^2(x)+\sin^2(x)]=\cos(x)\cdot1=\cos(x)\\\\L=\dfrac{\cos(x)}{\sin(x)}\cdot\sin(x)=\cos(x)\\\\L=R\\\\\text{used:}\\\\\sin^2x+\cos^2x=1\\\\\cot x=\dfrac{\cos x}{\sin x}$

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A sequence has an initial value (I) of -17 and its fourth term (a4) is 35. What is its general equation? Its 14th term?

aleksley [76]

Answer:

The sequence is an Arthemtic Progression

An=A1+(n-1)d

A1 is first term, An is nth term, n is number of term, and d is common difference

therefore

A4=35, A1= -17

A4=A1+(4-1)d

35= -17+3d

35+17=3d

52=3d

52/3=3d/3

14=d

common diffrence(d)=14

The general solution is given by

An= -17+(n-1)14

An= -17+14n-14

An= -31+14n

An= 14n-31

A14 term, means n=14

From An=A1+(n-1)d

A14= -17+(14-1)14

= -17+(13×14)

= -17+182

= 165.

Therfore, the 14th term is 165.

2. A sequence has a CR of 4/5 and its eighth term (a8) is (393216/3125). What is its general equation? Its 3rd term?

solution

common ratio(r)=4/5

eighth term(G8)=393216/3125

From Gn= G1r^(n-1)

G8 means n=8

G8=G1r^(n-1)

393216/3125=G1(4/5)^(8-1)

393216/3125=G1(4/5)^7

G1=(393216/3125)/(4/5)^7

G1=600

The first term is given by G1=600

Therefore

The General equation is given by

The General equation is given by Gn= 600(4/5)^(n-1)

3rd term (G3)

G3= G1(4/5)^(3-1) where n=3,

=600(4/5)^2

=600(16/25)

=384

Therefore, the 3rd term is given by G3= 384.

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3 years ago

Using the discriminatory, determine the number of real solution 2x^2+7x-15=0

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Answer:

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Step-by-step explanation:

x = -5 , x = 1.5

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3 years ago

for a March past competition on sports day the house captain of one house cut 25 pieces of ribbon each 1/2 metre in length.what

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Answer:

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Step-by-step explanation:

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3 years ago

Solve the equation X=

4vir4ik [10]

Step-by-step explanation:

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2 years ago

Perform the following operations and write the answers in radical form. Part A:√7+√3+√98−√18 Part B:3√5−3√11+2√121−3√90

Sveta_85 [38]

Answer:

$\sqrt{7}+\sqrt{3}+4\sqrt{2}$
$3\sqrt{5}-3\sqrt{11}+22-9\sqrt{10}$

Step-by-step explanation:

Part A ;

$\sqrt{7} +\sqrt{3} +\sqrt{98} -\sqrt{18} \\\\\sqrt{98}=7\sqrt{2}\\\sqrt{18}=3\sqrt{2}\\\\=\sqrt{7}+\sqrt{3}+7\sqrt{2}-3\sqrt{2}\\\\\mathrm{Add\:similar\:elements:}\:7\sqrt{2}-3\sqrt{2}=4\sqrt{2}\\\\=\sqrt{7}+\sqrt{3}+4\sqrt{2}$

Part B ;

$3\sqrt{5} - 3\sqrt{11} + 2\sqrt{121} -3\sqrt{90} \\\\2\sqrt{121}=22\\3\sqrt{90}=9\sqrt{10}\\\\=3\sqrt{5}-3\sqrt{11}+22-9\sqrt{10}$

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3 years ago