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Aliun [14]
3 years ago
8

Michael exchanged 1,000 US dollars for the Croation currency which is called Kuna. The exchange rate was 5.81 dollars to 1 dolla

r. How much croation kuna did Michael get?
Mathematics
1 answer:
steposvetlana [31]3 years ago
3 0
He can get 6,524.42 kuna
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Fill in the blank.<br> 65+26 is the same as 61 +
solniwko [45]

Answer:

30

Step-by-step explanation:

3 0
4 years ago
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Quadrilateral LMNO has diagonals that intersect at point P. If LP=NP, MP = y + 25, and OP = 5y + 29, find the length of MO such
cupoosta [38]
I saw the image of this problem.

For the quadrilateral to be parallelogram, MP = PO

MP = y + 25
PO = 5y + 29

y + 25 = 5y + 29
y - 5y = 29 - 25
-4y = 4
y = 4/-4
y = -1

MO = MP + OP
MO = y + 25 + 5y + 29
MO = -1 + 25 + 5(-1) + 29
MO = -1 + 25 - 5 + 29
MO = -6 + 54
MO = 48  Choice B.
8 0
4 years ago
(14x-5)-(-3x+2)<br><br>A. 11x-3<br>B. 17x-7<br>C. 17x-3<br>D. 11x-7
Vladimir [108]
The answer would be B) 17x-7
4 0
3 years ago
Find the equation of a line with a slope of 1 and passing through (11;3)
romanna [79]

Step-by-step explanation:

slope(m)=(Y-Y1)/X-X1

here,

M=1

(X1,Y1)=(11,3)

NOW,

m=(Y-Y1)/(X-X1)

1=(Y-3)/(X-11)

1×(X-11)=Y-3

X-11=Y-3

X-Y-11+3=0

X-Y-8=0

X-Y=0 is the required equation.

3 0
3 years ago
Help me Please....................
MakcuM [25]

9514 1404 393

Answer:

  1.3363

Step-by-step explanation:

The basic idea here is to find an expression for the direction vector between a point on L1 and a point on L2. Then, solve for the points on L1 and L2 that make that vector perpendicular to both lines L1 and L2. (The dot product of direction vectors is zero.) The distance between the points found is the shortest distance between the lines.

__

Let P be a point on L1. Then the parametric equation for P is ...

  P = (6t, 0, -t) . . . . . . origin + t × direction vector

Let Q be a point on L2. The direction vector for L2 is given by the difference between the given points. It is (4-1, 1-(-1), 6-1) = (3, 2, 5). Then the parametric equation for Q is ...

  Q = (3s+1, 2s-1, 5s+1) . . . . (1, -1, 1) + s × direction vector

The direction vector for PQ is ...

  Q -P = (3s+1-6t, 2s-1, 5s+1+t)

The dot product of this and the two lines' direction vectors will be zero:

  (3s+1-6t, 2s-1, 5s+1+t)·(6, 0, -1) = 0 = 13s -37t +5 . . . perpendicular to L1

  (3s+1-6t, 2s-1, 5s+1+t)·(3, 2, 5) = 0 = 38s -13t +6 . . . perpendicular to L2

The solution to these equations is ...

  s = -157/1237

  t = 112/1237

Then (Q-P) becomes (94, -1551, 564)/1237, and its length is ...

  |PQ| = √(94² +1551² +564²)/1237 ≈ 1.3363

The distance between the two lines is about 1.3363 units.

8 0
3 years ago
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