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frozen [14]
3 years ago
7

What length is Idaho

Mathematics
2 answers:
umka21 [38]3 years ago
8 0
Ccccccccccccccccccccc
MariettaO [177]3 years ago
7 0

Answer:

the length is 479 miles (771 km)

Step-by-step explanation:

the width is 305 miles (491 km)

the elevation is 5,000 ft (1,520 meters)

The highest elevation 12,662 ft ( borah peek) (3859 meters)

I hope this helps a lot and don,t believe me you can look it up  :)

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In the diagram, AC is a diameter of the circle with center 0. If m ZACB = 50°, what is ZBAC?
Snowcat [4.5K]

Answer:

40degrees

Step-by-step explanation:

Find the diagram attached

Since the triangle ina semi circle is a right angle, hence <B = 90degrees

Also frm the diagram

<A + <B + <C = 180

<A + 90 + 50 = 180

<A + 140 = 180

<A  = 180 - 140

<A = 40degrees

Hence the measure if m<BAC is 40degrees

7 0
3 years ago
Fran bought some art supplies from an online retailer. The retailer charges a $5.95 shipping fee or 6% of the total purchase pri
rodikova [14]

Answer:

Step-by-step explanation:

if 124.50 represents  100%

how much x  represents 6% ?

cross multiply

x= 124.50*6/100= 7.47

$7.47 > $5.95 so it will pay $7.47 in shiping

4 0
3 years ago
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PLZ HELP I GIVE BRAINLIEST!!! (Plz show work)
Triss [41]

Answer:

800 × ( 3.87 ÷ 100 ) × 1

= 30.96

3 0
3 years ago
Falling object formula: h=-16t+s
vitfil [10]

Answer:

  • 34.714 seconds

Step-by-step explanation:

<u>According to the question we have:</u>

  • Formula h = - 16t + s
  • h = 0
  • s = 555.427 ft
  • t = ?

<u>Substitute the given values and solve for t:</u>

  • 0 = - 16t + 555.427
  • 16t = 555.427
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3 0
2 years ago
Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S
Sophie [7]

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 A=J^{-1}AJ. 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 A=P^{-1}BP. In this equality we can perform a right multiplication by P^{-1} and obtain AP^{-1} =P^{-1}B. Then, in the obtained equality we perform a left multiplication by P and get PAP^{-1} =B. If we write Q=P^{-1} and Q^{-1} = P we have B = Q^{-1}AQ. 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 A=P^{-1}BP and from B↔C we have B=Q^{-1}CQ. Now, if we substitute the last equality into the first one we get

A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP).

Recall that if P and Q are invertible, then QP is invertible and (QP)^{-1}=P^{-1}Q^{-1}. So, if we denote R=QP we obtained that

A=R^{-1}CR. Hence, A↔C.

Therefore, the relation is an <em>equivalence relation</em>.

4 0
3 years ago
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