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2 years ago

Factor completely 81x2 − 100. (1 point) (10x − 9)(10x − 9) (10x − 9)(10x + 9) (9x − 10)(9x + 10) (9x − 10)(9x − 10)

Mathematics

1 answer:

MArishka [77]2 years ago

5 0

The complete factorization of the equation 81x² - 100 is; (9x - 10)(9x + 10)

<h3>How to factorize quadratic equations?</h3>

We are given the quadratic equation;

81x² - 100

Now, according to quadratic identities, we know that;

(a + b) * (a - b) = a² - b²

Now, our equation can also be expressed as;

81x² - 100 = 9²x² - 10²

Thus, applying the quadratic identity gives us;

(9x + 10)(9x - 10)

Read more about factorization of quadratic equations at; brainly.com/question/1214333

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3 years ago

Help me Please....................

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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.