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ad-work [718]
3 years ago
9

A pole that is 2.5 M tall cast a shadow that is 1.72M lawn dart at the same time a nearby tower cast a shadow that is 50.5 M lo

ng how tall is the tower round answer to the nearest meter
Mathematics
1 answer:
inn [45]3 years ago
5 0

Answer:

The tower is 73.4 m tall

Step-by-step explanation:

The height of the pole = 2.5 m

The shadow cast by the pole = 1.72 m

Shadow cast by tower = 50.5 m

To find the height of the tower, we proceed by finding the angle of elevation, θ, of the light source casting the shadows as follows;

Tan\theta =\dfrac{Opposite \ side \ to\  angle \ of \ elevation}{Adjacent\ side \ to\  angle \ of \ elevation} = \dfrac{Height \ of \ pole }{Length \ of \ shadow} =\dfrac{2.5 }{1.72}

\theta = tan ^{-1} \left (\dfrac{2.5 }{1.72} \right) = 55.47 ^{\circ}

The same tanθ gives;

Tan\theta = \dfrac{Height \ of \ tower}{Length \ of \ tower \  shadow} =\dfrac{Height \ of \ tower }{50.5} = \dfrac{2.5}{1.72}

Which gives;

{Height \ of \ tower } = {50.5} \times  \dfrac{2.5}{1.72} = 73.4 \ m

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Part B: what best describes the shape of the two graphs? Answer: 5th grade is roughly symmetrical without letters in fourth grade is skewed right

Step-by-step explanation: That’s for the answers for the 3rd question on Topic 4 of Section 9. Hope that helped:))

3 0
3 years ago
Leon's older brother is 4 3/4 feet tall. Leon is 1/3 foot shorter than his brother. How tall is Leon?
galben [10]
Wouldnt it be 4 2/4 feet tall?
3 0
3 years ago
given examples of relations that have the following properties 1) relexive in some set A and symmetric but not transitive 2) equ
rodikova [14]

Answer: 1) R = {(a, a), (а,b), (b, a), (b, b), (с, с), (b, с), (с, b)}.

It is clearly not transitive since (a, b) ∈ R and (b, c) ∈ R whilst (a, c) ¢ R. On the other hand, it is reflexive since (x, x) ∈ R for all cases of x: x = a, x = b, and x = c. Likewise, it is symmetric since (а, b) ∈ R and (b, а) ∈ R and (b, с) ∈ R and (c, b) ∈ R.

2) Let S=Z and define R = {(x,y) |x and y have the same parity}

i.e., x and y are either both even or both odd.

The parity relation is an equivalence relation.

a. For any x ∈ Z, x has the same parity as itself, so (x,x) ∈ R.

b. If (x,y) ∈ R, x and y have the same parity, so (y,x) ∈ R.

c. If (x.y) ∈ R, and (y,z) ∈ R, then x and z have the same parity as y, so they have the same parity as each other (if y is odd, both x and z are odd; if y is even, both x and z are even), thus (x,z)∈ R.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial but not transitive, so the relation provided in (1) satisfies this condition.

Step-by-step explanation:

1) By definition,

a) R, a relation in a set X, is reflexive if and only if ∀x∈X, xRx ---> xRx.

That is, x works at the same place of x.

b) R is symmetric if and only if ∀x,y ∈ X, xRy ---> yRx

That is if x works at the same place y, then y works at the same place for x.

c) R is transitive if and only if ∀x,y,z ∈ X, xRy∧yRz ---> xRz

That is, if x works at the same place for y and y works at the same place for z, then x works at the same place for z.

2) An equivalence relation on a set S, is a relation on S which is reflexive, symmetric and transitive.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial and not transitive.

QED!

6 0
3 years ago
A semi-circle window can be used above a doorway or as an accent window
anzhelika [568]

A semicircle is a part of a circle, and it is referred to as half of a given circle. Thud the area of the <em>semi-circle</em> window is 982 in^{2}.

A circle is a shape that is <u>bounded</u> by a <em>curved</em> path which is referred to as the <em>circumference</em>. Some <u>parts</u> of a circle are radius, diameter, sector, arc, semi-circle, circumference, etc.

A <em>semicircle</em> is a <u>part</u> of a <u>circle</u>, and it is referred to as <em>half </em>of a given <em>circle</em>.

such that:

<em>Area</em> of a <u>circle</u> = \pi r^{2}

and

area of a <u>semicircle</u> = \frac{\pi r^{2} }{2}

where: r is the <u>radius </u>of the <u>circle</u>, and \pi is a <u>constant </u>with a value of \frac{22}{7}.

Thus from the given question, it can be inferred that;

r = \frac{50}{2}

 = 25

r = 25 in

Thus, the area of the<em> semi-circle</em> can be determined as;

area of the <em>semi-circle</em> = \frac{1}{2} * \frac{22}{7} * 25^{2}

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area of the semi-circle = 982.14 in^{2}

The area of the <em>semi-circle</em> window is approximately 982 in^{2}.

for more clarifications on the area of a semi-circle, visit: brainly.com/question/15937849

#SPJ 1

8 0
2 years ago
Which of the following is true for an isometry?
FrozenT [24]
Isometry is defined as the <span> transformation that is invariant with respect to  the distance
The statement that is true for isometry that is corrrect is
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</span>so correct option is A as both the preimage and image are congruent
hope it helps<span />
5 0
3 years ago
Read 2 more answers
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