All categories

3 years ago

How is 3 help me please

Mathematics

1 answer:

Dovator [93]3 years ago

8 0

Step-by-step explanation:

Square root 8 = 2 square root 2

You might be interested in

PLEASE HELP ME GIVING 30 POINTS!!!

kramer

Using trigonometric identities, it is found that the sine and the tangent of the angle are given as follows:

$\sin{\theta} = \frac{1}{2}$ .

$\tan{\theta} = -\frac{\sqrt{3}}{3}$

<h3>How do we find the sine of an angle given the cosine?</h3>

We use the following identity:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

In this problem, the cosine is:

$\cos{\theta} = -\frac{\sqrt{3}}{2}$

Hence the sine is found as follows:

$\sin^{2}{\theta} + \left(-\frac{\sqrt{3}}{2}\right)^2 = 1$

$\sin^{2}{\theta} + \frac{3}{4} = 1$

$\sin^{2}{\theta} = \frac{1}{4}$

$\sin{\theta} = \pm \sqrt{\frac{1}{4}}$

Second quadrant, so the sine is positive, hence:

$\sin{\theta} = \frac{1}{2}$

<h3>What is the tangent of an angle?</h3>

The tangent is given by the <u>sine divided by the cosine</u>, hence:

$\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$

Hence:

$\tan{\theta} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

$\tan{\theta} = -\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$\tan{\theta} = -\frac{\sqrt{3}}{3}$

More can be learned about trigonometric identities at brainly.com/question/7331447

#SPJ1

5 0

1 year ago

Two pieces of fabric

Bess [88]

Answer:

<h2> 487/8 inches </h2>

Step-by-step explanation:

$\left(\dfrac{143}{4}-\dfrac38\right)+\left(\dfrac{207}{8}-\dfrac38\right)=\dfrac{286}{8}-\dfrac38+\dfrac{204}{8}=\dfrac{283}{8}+\dfrac{204}{8}=\dfrac{487}{8}$

6 0

3 years ago

Find the equation of a line parallel to y=4x−4 that contains the point (−2,0). Write the equation in slope-intercept form.

Allushta [10]

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above

$y=\stackrel{\stackrel{m}{\downarrow }}{4} x-4\qquad \impliedby \qquad \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so we're really looking for the equation of a line whose slope is 4 and passes through (-2 , 0)

$(\stackrel{x_1}{-2}~,~\stackrel{y_1}{0})\qquad \qquad \stackrel{slope}{m}\implies 4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{4}(x-\stackrel{x_1}{(-2)}) \\\\\\ y=4(x+2)\implies y=4x+8$