Ask question

Ask question

All categories

3 years ago

11

What is the height of the Tree? 2 m 3 m 30 m

1 answer:

lisabon 2012 [21]3 years ago

8 0

Answer:

The height of the tree is 3m

You might be interested in

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:

<u><em>Step(i):-</em></u>

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

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

<u><em>Step(i):-</em></u>

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

Calculus help! Using the mean value theorem!

marta [7]

$f$ is differentiable across its domain, so the MVT says there is some value of $c$ in the open interval $(a,b)$ such that

$f'(c)=\dfrac{f(b)-f(a)}{b-a}$

You have

$f(x)=Ax^2+Bx+C\implies f'(x)=2Ax+B$

so the equation above becomes

$2Ac+B=\dfrac{(Ab^2+Bb+C)-(Aa^2+Ba+C)}{b-a}$

Solve for $c$ .

$2Ac+B=\dfrac{A(b^2-a^2)+B(b-a)}{b-a}$
$2Ac+B=A(b+a)+B$
$2Ac=A(b+a)$
$c=\dfrac{b+a}2$

so $c$ is indeed the average of the endpoints, i.e. the midpoint.

8 0

3 years ago

The duration of routine operations at chicago general hospital has approximately a normal distribution with an average of 130 mi

vredina [299]

Answer:

79.77%

Step-by-step explanation:

P(X≥120) = normalcdf(120,1e99,130,12) = 0.7976716754 ≈ 0.7977 = 79.77%

Therefore, about 79.77% of operations last at least 120 minutes

5 0

3 years ago

An art class costs $45 for materials and then

Rudiy27

45+10x the x stands for how many classes you take

7 0

3 years ago

The graph represents this piecewise function

IrinaK [193]

Top one is x+3, bottom one is 5

6 0

3 years ago

Read 2 more answers