The ratio of red cars to blue cars in a parking lot was 7 : 2.If there were 35 red cars, how many blue cars were there?

MATH HELP PLZ!!!

RoseWind [281]

Answer:

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

b)

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

c)

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

d)

cos 330° = 1- 2 sin² (165°)

Step-by-step explanation:

Step(i):-

By using trigonometry formulas

a)

cos2∝ = 2 cos² ∝-1

cos∝ = 2 cos² ∝/2 -1

1+ cos∝ = 2 cos² ∝/2

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

b)

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

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

b)

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

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

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

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

d)

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

A boat sails on a bearing of 038°anf then 5km on a bearing of 067°.

I am Lyosha [343]

This question is not complete

Complete Question

A boat sails 4km on a bearing of 038 degree and then 5km on a bearing of 067 degree.(a)how far is the boat from its starting point.(b) calculate the bearing of the boat from its starting point

Answer:

a)8.717km

b) 54.146°

Step-by-step explanation:

(a)how far is the boat from its starting point.

We solve this question using resultant vectors

= (Rcos θ, Rsinθ + Rcos θ, Rsinθ)

Where

Rcos θ = x

Rsinθ = y

= (4cos38,4sin38) + (5cos67,5sin67)

= (3.152, 2.4626) + (1.9536, 4.6025)

= (5.1056, 7.065)

x = 5.1056

y = 7.065

Distance = √x² + y²

= √(5.1056²+ 7.065²)

= √75.98137636

= √8.7167296826

Approximately = 8.717 km

Therefore, the boat is 8.717km its starting point.

(b)calculate the bearing of the boat from its starting point.

The bearing of the boat is calculated using

tan θ = y/x

tan θ = 7.065/5.1056

θ = arc tan (7.065/5.1056)

= 54.145828196°

θ ≈ 54.146°

7 0

2 years ago

Use the definition of continuity to determine whether f is continuous at a. f(x) = 5x+5 a = -5 Question

Andrej [43]

ANSWER

$lim_{x \to - 5}(f(x)) = f( - 5)$

EXPLANATION

If f(x) is continuous at

$x = a$

Then,

$lim_{x \to a}(f(x)) = f(a)$

The given function is

$f(x) = 5x + 5$

$f( - 5) = 5( - 5) + 5$

$f( - 5) = - 25 + 5 = - 20$

$lim_{x \to - 5}(f(x)) = 5( - 5) + 5$

$lim_{x \to - 5}(f(x)) = - 20$

Since,

$lim_{x \to - 5}(f(x)) = f( - 5)$

The function is continuous at

$x = - 5$

4 0

3 years ago

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