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rjkz [21]
1 year ago
13

Two buildings on level ground are 120 m and 85 m tall respectively. given that the angle of elevation of the top of the taller b

uilding from the top of the shorter building is 33.9, find the distance between the two buildings
NOTE: this question is from trigonometry
Mathematics
1 answer:
Luda [366]1 year ago
7 0

Answer:

(-8,-5)

Step-by-step explanation:

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A certain firm has plants a, b, and c producing respectively 35\%, 15\%, and 50\% of the total output. The probabilities of a no
TEA [102]

The proportion of production that is defective and from plant A is

... 0.35·0.25 = 0.0875

The proportion of production that is defective and from plant B is

... 0.15·0.05 = 0.0075

The proportion of production that is defective and from plant C is

... 0.50·0.15 = 0.075

Thus, the proportion of defective product that is from plant C is

... 0.075/(0.0875 +0.0075 +0.075) = 75/170 = 15/34 ≈ 44.12%

_____

P(C | defective) = P(C&defective)/P(defective)

8 0
1 year ago
You receive an email asking you to forward it to four other people to ensure prosperity. Assuming the chain isn't broken, how ma
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well if it asks you to send to 4 people and there are 8 generations which includes yours that mean


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4 send to 4 each = 16 recived + the 4 before = 20 (generation 2)

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4 0
1 year ago
Read 2 more answers
If DE = 4x + 10, EF = 2x-1, and DF = 9x-15, find DF
Dmitrij [34]
Hello,
<span>We have that:
(4x+10)+(2x-1)=9x-15

</span>We solve the equation:
4x+10+2x-1=9x-15;
6x+9=9x-15;
6x-9x=-15-9;
-3x=-24;
3x=24;

<span>from which
</span>x=24:3=8

<span>Then:
DF=9x-15=(9</span>×8)-15=72-15=57

bye :-)
5 0
2 years ago
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Given: FH ⊥ GH; KJ ⊥ GJ Prove: ΔFHG ~ ΔKJG Identify the steps that complete the proof. ♣ = ♦ = ♠ =
Feliz [49]

1 is all right angles are congruent 2 angles fgh is congruent to angle kgj 3 aa similarity theorem just took the test this are correct

6 0
2 years ago
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Now imagine that instead of walking along the path 1→2→3→4→1, ann walks 80 meters on a straight line 33∘ north of east starting
stealth61 [152]

Answer:

Anna's walk  as a vector representation is 80\cos 33^{\circ}\hat{i}+80 \sin33^{\circ}\hat{j} and refer attachment.

Step-by-step explanation:

Let the origin be the point 1 from where Ann start walking.

Ann walks 80 meters on a straight line 33° north of the east starting at point 1 as shown in figure below,

Resolving into the vectors, the vertical component will be 80Sin33° and Horizontal component will be 80Cos33° as shown in figure (2)

Ann walk as a vector representation is 80\cos 33^{\circ}\hat{i}+80 \sin33^{\circ}\hat{j}

Thus, Anna's walk  as a vector representation is 80\cos 33^{\circ}\hat{i}+80 \sin33^{\circ}\hat{j}

 




7 0
1 year ago
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