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Law Incorporation [45]
2 years ago
12

Naomi says that the graph represnts a proportional relationship because it is a straight line. Is Naomi reasoning correct explai

n your answer.]
I need the answer rn please Brainly helmp me. Or someone I need this answer please.
Mathematics
1 answer:
Ahat [919]2 years ago
7 0

Answer:

sevensgyhiojdefhijfebufdk get biihrhd

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Determine whether the parabola y = −x2 + 15x + 8 opens up, down, left, or right.
spayn [35]
Dy/dx=-2x+15

d2y/dx2=-2

Since the acceleration is a constant negative it has an absolute maximum value for y(x), thus it opens downward, as x decreases without bound as x approaches ±∞
7 0
3 years ago
What is the difference between the absolute value of 4 and the absolute value of –3?
ra1l [238]

|4| - |-3| = 4 - 3 = 1 (answer)

6 0
3 years ago
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What is the slope of the line y=3x/7 + 4/7
Natalija [7]

Answer:

the slope of y=3x/7+4/7 is M=3 over 7

5 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
2 years ago
What is 1×3+45÷1×0=​
QveST [7]
The answer to this equation would be 3.
7 0
3 years ago
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