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ioda
2 years ago
12

Simplify these expressions:5m^2 x 2m60cf/3f15c-5f/2

Mathematics
1 answer:
Mars2501 [29]2 years ago
6 0
5m^2 x 2m = 10m^3
60cf/3f = 20cf^2
15c-5f/2 = Cannot be simplified further
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What is -1 11/12 in it’s simplest form
Dafna1 [17]

Answer: - 23/12

decimal -1.916

Step-by-step explanation:

3 0
3 years ago
A total of 628 tickets were sold for the school play. They were either adult tickets or student tickets. There were 72 fewer stu
myrzilka [38]
350 adult tickets were sold

4 0
3 years ago
2+d=2-3(d-5)-2 Pls show work
defon
2 + d = 2 - 3(d- 5) - 2

First, cancel 2 on both sides. / Your problem should look like: d = -3(d - 5) - 2
Second, expand. / Your problem should look like: d = -3d + 15 - 2
Third, simplify -3d + 15 - 2 to get -3d + 13. / Your problem should look like: d = -3d + 13
Fourth, add 3d to both sides. / Your problem should look like: d + 3d = 13
Fifth, add d + 3d to get 4d. / Your problem should look like: 4d = 13
Sixth, divide both sides by 4. / Your problem should look like: d =  \frac{13}{4}

Answer as fraction: d =  \frac{13}{4}
Answer as decimal: d = 3.25

3 0
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
I only have 10 minutes the answer the question please help me I will give you brainlist
Likurg_2 [28]
1 white shirt
1 black shirt
1 blue shirt
1 blue pant
1 tan pant
There are 5 total articles of clothing and 2 that he could end up wearing (white shirt & tan pants)
so the probability equals 2/5



7 0
2 years ago
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