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2 years ago

Find the solutions to x² = 12.

Mathematics

1 answer:

Novosadov [1.4K]2 years ago

3 0

Answer:

$x = \pm 2\sqrt3$

Step-by-step explanation:

Hello!

Let's use the square root property to solve this question

Solve:

$x^2 = 12$
$\sqrt{x^2} = \sqrt{12}$
$x = \pm\sqrt{12}$
$x = \pm \sqrt{4 * 3}$
$x = \pm 2\sqrt3$

The answer is option A: x = ±2√3

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3 years ago

A line passes through the point (-1,-5) and has a slope of -5.

Anna71 [15]

<h3>hello!</h3>

We're given a point that the line intersects and its slope.

Let's use Point-slope:-

$\boxed{y-y1=m(x-x1)}$

y1 = -5

m=-5

x1=-1

$\boxed{y-(-5)=-5(x-(-1)}$

$\boxed{y+5=-5(x+1)}$

Convert to slope-intercept:-

$\boxed{y+5=-5x-5}$

$\boxed{y=-5x-5-5}$

Finally,

$\bigstar{\boxed{\pmb{y=-5x-10}}$

Hope everything is clear; if you need any clarification/explanation, kindly let me know, and I will comment and/or edit my answer :)

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2 years ago

A vector with magnitude 9 points in a direction 190 degrees counterclockwise from the positive x axis. Write in component form

Over [174]

Answer:

$\vec{v}= \text{ or } \approx$

Step-by-step explanation:

Component form of a vector is given by $\vec{v}=$ , where $i$ represents change in x-value and $j$ represents change in y-value. The magnitude of a vector is correlated the Pythagorean Theorem. For vector $\vec{v}=$ , the magnitude is $||v||=\sqrt{i^2+j^2$ .

190 degrees counterclockwise from the positive x-axis is 10 degrees below the negative x-axis. We can then draw a right triangle 10 degrees below the horizontal with one leg being $i$ , one leg being $j$ , and the hypotenuse of the triangle being the magnitude of the vector, which is given as 9.

In any right triangle, the sine/sin of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle.

Therefore, we have:

$\sin 10^{\circ}=\frac{j}{9},\\j=9\sin 10^{\circ}$

To find the other leg, $i$ , we can also use basic trigonometry for a right triangle. In right triangles only, the cosine/cos of an angle is equal to its adjacent side divided by the hypotenuse of the triangle. We get:

$\cos 10^{\circ}=\frac{i}{9},\\i=9\cos 10^{\circ}$