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2 years ago

Can anyone help me with this question?

Mathematics

1 answer:

Harrizon [31]2 years ago

8 0

Answer: Here you go

Explanation:

Raduis is 2.5cm

Diameter 5cm

Circumference 15.70796cm or 16 rounded up because 2 times pie time 2.5 because 2.5 is the radius which if you calculate is 15.7 because 2 x 3.1415 x 2.5 = 15.7

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2 years ago

Liliana and Marcus are playing a word game. Liliana spells a word incorrectly and loses 18 points. Then Marcus spells a word cor

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2 years ago

Integrate cosx/sqrt(1+cosx)dx

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<span>Take the integral: integral (cos(x))/sqrt(cos(x)+1) dx
For the integrand (cos(x))/sqrt(1+cos(x)), substitute u = 1+cos(x) and du = -sin(x) dx: = integral (u-1)/(sqrt(2-u) u) du
For the integrand (-1+u)/(sqrt(2-u) u), substitute s = sqrt(2-u) and ds = -1/(2 sqrt(2-u)) du: = integral -(2 (1-s^2))/(2-s^2) ds
Factor out constants: = -2 integral (1-s^2)/(2-s^2) ds
For the integrand (1-s^2)/(2-s^2), cancel common terms in the numerator and denominator: = -2 integral (s^2-1)/(s^2-2) ds
For the integrand (-1+s^2)/(-2+s^2), do long division: = -2 integral (1/(s^2-2)+1) ds
Integrate the sum term by term: = -2 integral 1/(s^2-2) ds-2 integral 1 ds
Factor -2 from the denominator: = -2 integral -1/(2 (1-s^2/2)) ds-2 integral 1 ds
Factor out constants: = integral 1/(1-s^2/2) ds-2 integral 1 ds
For the integrand 1/(1-s^2/2), substitute p = s/sqrt(2) and dp = 1/sqrt(2) ds: = sqrt(2) integral 1/(1-p^2) dp-2 integral 1 ds
The integral of 1/(1-p^2) is tanh^(-1)(p): = sqrt(2) tanh^(-1)(p)-2 integral 1 ds
The integral of 1 is s: = sqrt(2) tanh^(-1)(p)-2 s+constant
Substitute back for p = s/sqrt(2): = sqrt(2) tanh^(-1)(s/sqrt(2))-2 s+constant
Substitute back for s = sqrt(2-u): = sqrt(2) tanh^(-1)(sqrt(1-u/2))-2 sqrt(2-u)+constant
Substitute back for u = 1+cos(x): = sqrt(2) tanh^(-1)(sqrt(sin^2(x/2)))-2 sqrt(1-cos(x))+constant
Factor the answer a different way: = sqrt(1-cos(x)) (csc(x/2) tanh^(-1)(sin(x/2))-2)+constant
Which is equivalent for restricted x values to:
Answer: | | = (2 cos(x/2) (2 sin(x/2)+log(cos(x/4)-sin(x/4))-log(sin(x/4)+cos(x/4))))/sqrt(cos(x)+1)+constant</span>

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3 years ago

A flagpole sits on the top of a cliff that is 25 m above sea level. If the angle of elevation from a boat on the water to the to

OLEGan [10]

Answer:

3.2 m

Step-by-step explanation:

Let x be the distance between the cliff and the boat.

Since the cliff is 25 m above sea level and the distance between the boat and the height of the cliff are perpendicular, and the line of sight of the boat to the base of a cliff is 29 , these three form a right-angled triangle with hypotenuse side as the line of sight of the elevation of the boat to the base of the cliff.

So, using trigonometric ratios, tan29° = 25/x

x = 25 m/tan29° = 25 m/0.5543 = 45.1 m

Also, the line of sight of the top of the flagpole, the cliff and the distance between the boat and the cliff form a right-angled triangle with hypotenuse side as the line of sight of the top of the flagpole,

Now, the angle of elevation of the top of the flagpole from the top of the boat is 32°, we now find the angle opposite to x, the distance from the boat to the cliff.

Since it is a right-angled triangle, thus one angle is 90° and the unknown angle is A.

So, 90° + 32° + A = 180° (sum of angles in a triangle)

122° + A = 180°

A = 180° - 122°

A = 58°

Let y be the length of the flagpole. In this same triangle, using the sine rule, with x being the side opposite A and 25 + y being the side opposite the 32° angle, where 25 + y = height of cliff + length of flagpole, then

x/sinA = (25 + y)/sin32°

25 + y = xsin32°/sinA

y = xsin32°/sinA - 25

y = 45.1sin32°/sin58° - 25

y = 45.1 × 0.5299/0.8480 - 25

y = 45.1 × 0.6249 - 25

y =28.183 - 25

y = 3.183

y ≅ 3.2 m to one decimal place

6 0

2 years ago