Find the prime factorization for 96

How many x-intercepts does the graph of each quadratic equation have? Show your work.

Alex73 [517]

Answer:

Step-by-step explanation:

In each case we find the discriminant b^2 - 4ac.

If the discriminant is negative, we have two unequal, complex roots.

If the discriminant is zero. we have two equal, real roots.

If the discriminant is positive, we have two unequal real roots.

#51: 8v^2 - 12v + 9: the discriminant is (-12)^2 - 4(8)(9) = -144. we have two unequal, complex roots

#52: (-11)^2 - 4(4)(-14) = 121 + 224 = 345. we have two unequal real roots.

#53: (-5)^2 - 4(7)(6) = 25 - 168 (negative). we have two unequal, complex roots.

#54: (4)^2 - 16 = 0. We have two equal, real roots.

4 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:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

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

How do i classify triangles?

12345 [234]

Answer:

By the angles and sides, but if you need more help here is a link to a video that could be pretty helpful.

Step-by-step explanation:

https://www.khanacademy.org/math/cc-fourth-grade-math/plane-figures/imp-classifying-triangles/v/scalene-isosceles-equilateral-acute-right-obtuse#:~:text=Learn%20to%20categorize%20triangles%20as,acute%2C%20right%2C%20or%20obtuse.

7 0

3 years ago

Read 2 more answers

The number of patients who reported that a new drug had relieved their pain. discrete or continuous?

solong [7]

Answer:

Continuously

Step-by-step explanation:

Continuously

4 0

3 years ago

Question 6: please help me out here :(((

Murljashka [212]

Answer:

m∠1 = 106

Step-by-step explanation:

Given: m∠5 = 106

Vertical angles are congruent

The Alternate Interior Angles Theorem states that, when two parallel lines are cut by a transversal , the resulting alternate interior angles are congruent.

Therefore: m∠1 = 106

7 0

2 years ago