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Vilka [71]
3 years ago
10

Help me with this questing plsss

Mathematics
1 answer:
docker41 [41]3 years ago
8 0

Answer:

6.25n + 3.50

$34.75

Step-by-step explanation:

Break down the important information given in the problem.

The one-time delivery fee is 3.50. This is only paid one time an never again, making it the <u>constant</u>, a number that does not change.

Each lunch costs 6.25. This amount will increase depending on how many times Mr. Jackson orders lunches, "n" times. This number is the <u>rate</u> because it changes. The rate is attached to the variable.

If you add the amounts together, that is the total cost of ordering lunches.

6.25n + 3.50

(Reember expressions do not have the equal sign).

To find the cost of ordering 5 lunches, use the expression. Substitute "n" for 5 because "n" represents the number of lunches ordered.

6.25n + 3.50

= 6.25(5) + 3.50   Simplify by multiplying 5 and 6.25

= 31.25 + 3.50    Add the two values

= 34.75       Total cost of 5 lunches

Therefore the cost of ordering 5 lunches is $34.75.

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Let AB be the line segment beginning at point A(0, 3) and ending at point B(6, -10). Find the point P on the line segment that i
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Answer:

Point P = 14.32

Step-by-step explanation:

Point P represents the distance from point A to point B.

The formula is given as:

P=\sqrt{(x_{2}-x_{1})  ^{2} + (y_{2}-y_{1})  ^{2}}

x1 = 0, y1 = 3, x2 = 6, and y2 = -10

P=\sqrt{(6-0)  ^{2} + (-10-3)  ^{2}}\sqrt{x} \\\\P = \sqrt{6^{2} + (-13)^{2} }= \sqrt{36+169} \\\\P = \sqrt{205}\\ \\P = 14.3178

P ≅ 14.32

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3 years ago
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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
(70 points) please answer this question and explain how you get the answer!!
musickatia [10]

Answer: -0.10

Step-by-step explanation:

there is a pattern because 2014 0.45, 2015 0.20, 2016 0.35, 2016 0.10.

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Answer:

A

Step-by-step explanation:

Remark

f and d are equal (and acute) because they are corresponding angles.

82 and f are supplementary, so we can find f

Finding f

f + 82 = 180

f = 180 - 82

f = 82

Finding d

d = f

d = 82

5 0
2 years ago
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