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o-na [289]
3 years ago
9

A map has a scale of 1 centimeter = 40 kilometer. If the distance between Fort Worth and El Paso is 21.7 cm on a map , how many

kilometers are between the two cities?
Mathematics
1 answer:
ryzh [129]3 years ago
5 0
Aloha!
 1 centimeter= 40 kilometers
The distance between the cities in centimeters is 21.7
 To get the answer into kilometers, you multiply.
So, 21.7 times 40 which is 868 kilometers.  
 To answer your question, the distance between Fort Worth and El Paso is 868 kilometers.
 I hope this helps!
  Adios!:)
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If the two equations in parenthesizes were there own equations which ones would equal 1 and 6
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3 years ago
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Py+7=6y+qp, y, plus, 7, equals, 6, y, plus, q y=y=y, equals
Archy [21]

Py +7 = 6y + qp

Solve the equation for y

To solve the equation for y we need to get y alone

Py +7 = 6y + qp

Subtract 6y from both sides

Py - 6y +7 = + qp

Subtract 7 from both sides

Py - 6y = + qp - 7

Now factor out y

(P - 6)y = qp - 7

Divide by P - 6 from both sides

y =\frac{qp - 7}{(P - 6)}

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mafiozo [28]

Answer:

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Step-by-step explanation:

it may be thought of as the "middle" value of a data set. For example, in the data set {1, 3, 3, 6, 7, 8, 9}, the median is 6,

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2 years ago
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In ΔABC (m∠C = 90°), the points D and E are the points where the angle bisectors of ∠A and ∠B intersect respectively sides BC an
Airida [17]

This is a little long, but it gets you there.

  1. ΔEBH ≅ ΔEBC . . . . HA theorem
  2. EH ≅ EC . . . . . . . . . CPCTC
  3. ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
  4. ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
  5. ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
  6. ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
  7. ΔDAC ≅ ΔDAG . . . HA theorem
  8. DC ≅ DG . . . . . . . . . CPCTC
  9. ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
  10. ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
  11. ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
  12. ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
  13. (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
  14. ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
  15. This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
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3 years ago
Is 1/2 greater than 4/5
Law Incorporation [45]
For this question you should know:
1/2 = 5/10 
and 
4/5 = 8/10 
so you can see 8/10 is greater than 5/10 so the answer is no 
4/5 is greater than 1/2 :)))
i hope this is helpful 
have a nice day 
4 0
3 years ago
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