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dexar [7]
3 years ago
10

What is the distance between 5 and -4

Mathematics
1 answer:
marta [7]3 years ago
7 0

Answer:

8

Step-by-step explanation:

you just count backwards from 5 until you reach -4

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Correct answer will get brainliest.
ehidna [41]

Answer:576

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Need a little help with one please
Pani-rosa [81]

Answer:

That would be the side-side-side (SSS) postulate, which states that if all the sides of a triangle are in a fixed ratio to all the corresponding sides of another triangle, then the two triangles are said to be congruent.

Step-by-step explanation:

Looking at triangles ABC and DEF, we notice that:

\frac{AB}{DE} = \frac{AC}{DF}=\frac{BC}{EF}

since

\frac{3}{18} = \frac{6}{36}=\frac{7}{42}=\frac{1}{6}

Let me know if you have further questions.

4 0
3 years ago
What is the y-intercept of the graph that is shown below?
Gnoma [55]

Answer:

(0,4)

Step-by-step explanation:

The y intercept is where the x value is zero

x=0 and y =4

The y intercept is (0,4)

7 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
What is the value of x?
Makovka662 [10]

Answer:

A-100

Step-by-step explanation:

Use the Pythagorean therom.

a^2+b^2=c^2

6^2+8^2=c^2

36+64=100.

Hope this helps. Kinda confused me.


4 0
3 years ago
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