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2 years ago

Drew dis its one of kmk charactersssss

Mathematics

1 answer:

valentina_108 [34]2 years ago

5 0

Step-by-step explanation:

yo that's actually really sick

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Consider a triangle formed by connecting the three points: (0 0 0), (1 0 0), (1 1 1). a. Find the vector which is perpendicular

Sidana [21]

Answer:

=k−j or (0 -1 1)

Step-by-step explanation:

The Vector or Outer Product two vectors, say →x&→y, is

denoted by →x×→y, is a perpendicular vector to the plane containing them.

The given pts. A (0 0 0), B (1 0 0), C (1 1 1) lie in a

plane ABC.

As suchType equation here., the vectors (→(AB ) )and (→AC ) are members, ∈ of the plane ABC

That is □(→(AB )×(→AC)) to the plane ABC

(→(AB ) )=(1-0,0-0,0-0)=(1,0,0)

(→(AC ) )=(1-0,1-0,1-0)=(1,1,1)

(→(AB )×(→AC= (0i -j k)

=k−j

Which is in the positive Z direction

3 0

2 years ago

Solve 3x+k= c for x.

Debora [2.8K]

Answer:

x = $\frac{c-k}{3}$

Step-by-step explanation:

Given

3x + k = c ( isolate the term 3x by subtracting k from both sides )

3x = c - k ( divide both sides by 3 )

x = $\frac{c-k}{3}$

5 0

2 years ago

What is the biggest decimal and take away two

tensa zangetsu [6.8K]

Answer:

about -1

Step-by-step explanation:

0.9999999999 minus 2

3 0

2 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$