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3 years ago

MATH

Mathematics

1 answer:

ale4655 [162]3 years ago

7 0

Answer:

the answer is 3

Step-by-step explanation:

in the number 66 there are "2" circles in them and so the answer we get 2

in the number 99 there are "2" circles and so the answer there is 2

in the number 888 there are "6" circles so the answer is 6

in the number 00 there are "2" circles that's the answer

in the number 7777 there are no circles so the answer is 0

we keep on going like that......

in the number 2876 there are 3 circles 2 in the number 8 and 1 in the number 6

hope this helped :D

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MATH HELP PLZ!!!

RoseWind [281]

Answer:

a) $tan (157.5) = \frac{1-cos 315}{sin315}$

$sin (165) =\sqrt{ \frac{1-cos (330) }{2}}$

$sin^{2} (157.5) = \frac{1-cos (315) }{2}$

cos 330° = 1- 2 sin² (165°)

Step-by-step explanation:

Step(i):-

By using trigonometry formulas

cos2∝ = 2 cos² ∝-1

cos∝ = 2 cos² ∝/2 -1

1+ cos∝ = 2 cos² ∝/2

$cos^{2} (\frac{\alpha }{2}) = \frac{1+cos\alpha }{2}$

cos2∝ = 1- 2 sin² ∝

cos∝ = 1- 2 sin² ∝/2

$sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}$

Step(i):-

Given

$tan\alpha = \frac{sin\alpha }{cos\alpha }$

we know that trigonometry formulas

$sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )$

1- cos∝ = 2 sin² ∝/2

Given

$tan(\frac{\alpha }{2} ) = \frac{sin(\frac{\alpha }{2} )}{cos(\frac{\alpha }{2}) }$

put ∝ = 315

$tan(\frac{315}{2} ) = \frac{sin(\frac{315 }{2} )}{cos(\frac{315 }{2}) }$

multiply with ' 2 sin (∝/2) both numerator and denominator

$tan (\frac{315}{2} )= \frac{2sin^{2}(\frac{315)}{2} }{2sin(\frac{315}{2} cos(\frac{315}{2}) }$

Apply formulas

$sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )$

1- cos∝ = 2 sin² ∝/2

now we get

$tan (157.5) = \frac{1-cos 315}{sin315}$

$sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}$

put ∝ = 330° above formula

$sin^{2} (\frac{330 }{2}) = \frac{1-cos (330) }{2}$

$sin^{2} (165) = \frac{1-cos (330) }{2}$

$sin (165) =\sqrt{ \frac{1-cos (330) }{2}}$

c )

$sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}$

put ∝ = 315° above formula

$sin^{2} (\frac{315 }{2}) = \frac{1-cos (315) }{2}$

$sin^{2} (157.5) = \frac{1-cos (315) }{2}$

cos∝ = 1- 2 sin² ∝/2

put ∝ = 330°

$cos 330 = 1 - 2sin^{2} (\frac{330}{2} )$

cos 330° = 1- 2 sin² (165°)

3 0

3 years ago

Find the first term and the common difference of the arithmetic sequence whose 5 th term is 3 and 19 th term is 45 .

gizmo_the_mogwai [7]

Where (a) is first term
and common diff is (d)

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3 years ago

Somebody please help!!!!!!!!

Jlenok [28]

Answer:

I think the answer is b^2=8^2-4^2

Step-by-step explanation:

The equation is A^2+B^2=C^2

C^2-A^2=B^2

C^2-B^2=A^2

C is the hypotenuse. You know what a is and all you have to do is show the equation

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nirvana33 [79]

The probability is 3/70

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What is the value of c in the equation below? StartFraction 5 Superscript 5 Baseline Over 5 squared EndFraction = a Superscript

maw [93]

Answer:

125

Step-by-step explanation:

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