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2 years ago

MARKING BRAINLEST!!! ALSO 5 STAR

Mathematics

1 answer:

Bogdan [553]2 years ago

5 0

Answer:

a: $6^{3}$ , 216 b: -1(3²), -9

Step-by-step explanation:

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Find Value of x for 6

Elis [28]

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2 years ago

ANSWER PLEASE!!!!! ASAP!!! THANKS!!

GREYUIT [131]

Answer:

The answer is 64.

Step-by-step explanation:

To find the volume of a prism, you multiply l x w x h (length x width x height). The length is 8, width is 4, height is 2. 8 x 4 = 32, 32 x 2 = 64.

hope this helped plz tell me if i was right!!!

8 0

2 years ago

What must be added to a^2+3/5a

kolezko [41]

This question doesn’t really make sense- is it multiple choice, if so what are the options? A can be anything as long as it’s the same?

3 0

2 years ago

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

2 years ago

Un pueblo tenía 13.578 habitantes en 1970. En 1988 la población se incrementó 1,5 veces con respecto a 1970 y en 2001 se increme

igor_vitrenko [27]

Answer:

En 2001 el pueblo poseía 45.826 habitantes.

Step-by-step explanation:

Un pueblo tenía 13.578 habitantes en 1970. En 1988 la población se incrementó 1,5 veces con respecto a 1970. Para calcular entonces la población en 1988 debes realizar el siguiente cálculo:

13.578*1,5= 20.367

Por lo que en 1988 el pueblo poseía 20.367 habitantes.

En 2001 se incrementó la población 2,25 veces en relación a 1988. Para calcular entonces la población en 2001 debes realizar el siguiente cálculo:

20.367* 2,25= 45.825,75≅ 45.826

Por lo que en 2001 el pueblo poseía 45.826 habitantes.

8 0

2 years ago