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jeyben [28]
2 years ago
7

Carl knitted 4 more scarves this year than last year. If he knitted 16% more scarves this year than last year, how many did he k

nit this year?
Mathematics
1 answer:
Readme [11.4K]2 years ago
7 0

Answer:25

Step-by-step explanation:

4 -> 16%

x->100%

x=400/16=25

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A, C, and E are the answers

Step-by-step explanation:

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2 years ago
JK⎯⎯⎯⎯⎯ is dilated by a scale factor of n with the origin as the center of dilation, resulting in the image J′K′⎯⎯⎯⎯⎯⎯⎯. The slo
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3 years ago
Read 2 more answers
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
Choose the most appropriate strategy and then use it to solve the quadratic equation.
dsp73

 (x-8) ^ 2 = 121
 
(x-8) = + / - root (121)
 x = 8 +/- root (121)
 The solutions are:
 
x1 = 8 + root (121)
 
x2 = 8 - root (121)
 2a ^ 2 = 8a-6
 2a ^ 2-8a + 6 = 0
 a ^ 2-4a + 3 = 0
 (a-1) (a-3) = 0
 The solutions are:
 
a1 = 1
 
a2 = 3
 x ^ 2 + 12x + 36 = 4
 x ^ 2 + 12x + 36-4 = 0
 x ^ 2 + 12x + 32 = 0
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x1 = -8
 
x2 = -4
 x ^ 2-x + 30 = 0
 
x = (- b +/- root (b ^ 2 - 4 * a * c)) / 2 * a
 x = (- (- 1) +/- root ((- 1) ^ 2 - 4 * (1) * (30))) / 2 * (1)
 x = (1 +/- root (1 - 120))) / 2
 x = (1 +/- root (-119))) / 2
 x = (1 +/- root (119) * i)) / 2
 The solutions are:
 
x1 = (1 + root (119) * i)) / 2
 
x2 = (1 - root (119) * i)) / 2
7 0
3 years ago
What are two other ways to name plane C?
svetoff [14.1K]
Answer: Other names for plane C can be plane EFG or plane EBF hope that helps (:
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2 years ago
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