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Solnce55 [7]
3 years ago
10

How do you factor x² + 7x + 12

Mathematics
1 answer:
wel3 years ago
5 0
If you mean: x squared -7x +12 then it is (x-3)(x-4) when factored .
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#10 using right angle below find the tangent of angle A.
just olya [345]

Answer: the first option is the correct answer.

Step-by-step explanation:

Triangle ABC is a right angle triangle.

From the given right angle triangle,

AB represents the hypotenuse of the right angle triangle.

With m∠A as the reference angle,

AC represents the adjacent side of the right angle triangle.

BC represents the opposite side of the right angle triangle.

To determine the tangent of angle A, we would apply the Tangent trigonometric ratio. It is expressed as

Tan θ, = opposite side/adjacent side. Therefore,

Tan A = 5/5√3 = 1/√3

Rationalizing the surd, it becomes

1/√3 × √3/√3

Tan A = √3/3

6 0
3 years ago
Find two vectors in R2 with Euclidian Norm 1<br> whoseEuclidian inner product with (3,1) is zero.
alina1380 [7]

Answer:

v_1=(\frac{1}{10},-\frac{3}{10})

v_2=(-\frac{1}{10},\frac{3}{10})

Step-by-step explanation:

First we define two generic vectors in our \mathbb{R}^2 space:

  1. v_1 = (x_1,y_1)
  2. v_2 = (x_2,y_2)

By definition we know that Euclidean norm on an 2-dimensional Euclidean space \mathbb{R}^2 is:

\left \| v \right \|= \sqrt{x^2+y^2}

Also we know that the inner product in \mathbb{R}^2 space is defined as:

v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

\left \| v_1 \right \|= \sqrt{x^2+y^2}=1

and

\left \| v_2 \right \|= \sqrt{x^2+y^2}=1

As second condition we have that:

v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0

v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0

Which is the same:

y_1=-3x_1\\y_2=-3x_2

Replacing the second condition on the first condition we have:

\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}

Since x_1^2= \frac{1}{10} we have two posible solutions, x_1=\frac{1}{10} or x_1=-\frac{1}{10}. If we choose x_1=\frac{1}{10}, we can choose next the other solution for x_2.

Remembering,

y_1=-3x_1\\y_2=-3x_2

The two vectors we are looking for are:

v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})

5 0
3 years ago
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