Answer: The answer is 381.85 feet.
Step-by-step explanation: Given that a window is 20 feet above the ground. From there, the angle of elevation to the top of a building across the street is 78°, and the angle of depression to the base of the same building is 15°. We are to calculate the height of the building across the street.
This situation is framed very nicely in the attached figure, where
BG = 20 feet, ∠AWB = 78°, ∠WAB = WBG = 15° and AH = height of the bulding across the street = ?
From the right-angled triangle WGB, we have
and from the right-angled triangle WAB, we have'
Therefore, AH = AB + BH = h + GB = 361.85+20 = 381.85 feet.
Thus, the height of the building across the street is 381.85 feet.
The total number of digits used when the first 2002 positive even integers are written is 4 digits
<h3>How to find the number of sequence?</h3>
Using arithmetic sequence,
nth term = a + (n - 1)d
where
- a = first term
- d = common difference
- n = number of terms
a = 2
d = 2
nth term = 2 + (2002 - 1)2
nth term = 2 + 4002
nth term = 4004
learn more on sequence here: brainly.com/question/15456604
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The answer would be -0.102564 repeating because if you do the math before dividing you would get -8/6.5/-1.5/-8
Answer:
SY = 16
XY = 12
Step-by-step explanation:
15/20 = 12/SY
15SY = 240
SY = 16
15/20 = 9/XY
15XY = 180
XY = 12
Answer:
Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.
<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that 
. Thus, A↔A.
<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that 
. In this equality we can perform a right multiplication by 
and obtain 
. Then, in the obtained equality we perform a left multiplication by P and get 
. If we write 
and 
we have 
. Thus, B↔A.
<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have and from B↔C we have 
. Now, if we substitute the last equality into the first one we get
.
Recall that if P and Q are invertible, then QP is invertible and 
. So, if we denote R=QP we obtained that
. Hence, A↔C.
Therefore, the relation is an <em>equivalence relation</em>.
