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3 years ago

Please help ASAP and explain the answer, thank you

Mathematics

1 answer:

jonny [76]3 years ago

6 0

Answer:

Step-by-step explanation:

1) x² - 8x + 5 = 0

2) No subtract 5 from both sides

x² - 8x + 5 - 5 = 0 - 5

x² - 8x = -5

3) Now find (8/2) = 4. Add 4² = 16 to both sides

x² - 8x + 16 = -5 + 16

x² - 8x + 16 = 11

4) (x - 4)² = 11

5) Take square root both sides

x - 4 = ±√11

x = 4 ±√11

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Is 59.05 greater than 59.5

bekas [8.4K]

No, 59.05 is smaller because you have to add 0.45 to get 59.5.

8 0

3 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

3 years ago

The semi circle and isosceles triangle have the same base AB and the same area. Find angle x. Show workout.

Nina [5.8K]

$r= \frac{AB}{2} \\ \\A_{semi\:circle }=\frac{r^2\pi}{2} \\ \\A_{triangle}=A_{semi\:circle }=\frac{r^2\pi}{2} \\ \\A_{triangle}= \frac{2r\times h}{2} =rh \\ \\\frac{r^2\pi}{2}= rh \\ \\h=\frac{r\pi}{2} \\ \\ \tan{x}= \frac{h}{r} = \frac{\frac{r\pi}{2}}{r} =\frac{\pi}{2} \\ \\x=\arctan{\frac{\pi}{2}}
\\
\\x \approx 57.52^o$