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Dahasolnce [82]
3 years ago
10

Set up a right triangle model for this problem and solve by using the reference table trigonometric ratio that applies. Follow t

he models above.
A photographer stands 60 yards from the base of a lighthouse and observes that the angle between the ground and the top of the lighthouse is 41°. How tall is the lighthouse?
A.45.3 yards
B.52.2 yards
C.39.4 yards
Mathematics
2 answers:
RSB [31]3 years ago
5 0
In reference to the angle of elevation (41 degrees), the adjacent side is 60 and the opposite side is the unknown.
Using tangent ratio:
tan 41 = opp/60
60 tan41 = opp
opp side = 52.2yards
LETTER B
g100num [7]3 years ago
5 0

Answer: B. 52.2 yards

Lighthouse is 52.2 yards tall.

Step-by-step explanation:

Let AB denotes the height of the lighthouse & BC denotes the distance between the base of a lighthouse and Photographer.

In Figure,  ∠ACB = 41° & ∠ABC = 90°, Threrefore,  Base is BC and perpendicular is AB.

As we know, tan(∠ACB) = \frac{Perpendicular}{Base}

∴  tan 41° = \frac{AB}{BC}

tan 41° = \frac{AB}{60}

0.869286737 = \frac{AB}{60}

AB = 52.2 yards ( approx.)

Thus, Height of the lighthouse is 52.2 yards


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Is the algebra, if so what type?
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Your friend incorrectly factors the expression 15x — 20xy as 5x(3— 4xy
nika2105 [10]

Answer:

no 15x — 20xy as 5x(3— 4xy is not correct

Step-by-step explanation:

15x — 20xy ≠ 5x(3— 4xy)

there are not equal because 5x(3 - 4xy) =

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6 0
3 years ago
A poll that does not attempt to generate a random sample, but instead invites people to volunteer to participate is called:_____
Helga [31]

Answer:

Self selection sampling.

Step-by-step explanation:

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6 0
3 years ago
your grades on tests 1 2 and 4 are 82 76 and 90. unfortunately you cut the third test and received a 0. If you have one test lef
Dmitrij [34]

Answer:  No, he can't pass the course.

Explanation:

Since we have given that

Marks on tests 1, 2, and 4 are given by

82, 76, 90 respectively.

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According to question, we have given that passing grade for the course is 70, and his one test left to take ,

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8 0
3 years ago
3.
loris [4]

(a) It looks like the ODE is

<em>y'</em> = 4<em>x</em> √(1 - <em>y </em>^2)

which is separable:

d<em>y</em>/d<em>x</em> = 4<em>x</em> √(1 - <em>y</em> ^2)   =>   d<em>y</em>/√(1 - <em>y</em> ^2) = 4<em>x</em> d<em>x</em>

Integrate both sides. On the left, substitute <em>y</em> = sin(<em>t </em>) and d<em>y</em> = cos(<em>t</em> ) d<em>t</em> :

∫ d<em>y</em>/√(1 - <em>y</em> ^2) = ∫ 4<em>x</em> d<em>x</em>

∫ cos(<em>t</em> ) / √(1 - sin^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

∫ cos(<em>t</em> ) / √(cos^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

∫ cos(<em>t</em> ) / |cos(<em>t</em> )| d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

Since we want the substitutiong to be reversible, we implicitly assume that -<em>π</em>/2 ≤ <em>t</em> ≤ <em>π</em>/2, for which cos(<em>t</em> ) > 0, and in turn |cos(<em>t</em> )| = cos(<em>t</em> ). So the left side reduces completely and we get

∫ d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

<em>t</em> = 2<em>x</em> ^2 + <em>C</em>

arcsin(<em>y</em>) = 2<em>x</em> ^2 + <em>C</em>

<em>y</em> = sin(2<em>x</em> ^2 + <em>C </em>)

(b) There is no solution for the initial value <em>y</em> (0) = 4 because sin is bounded between -1 and 1.

7 0
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