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2 years ago

Um. i just need help with this! so if anyone knows help me plz <3

Mathematics

2 answers:

Zinaida [17]2 years ago

7 0

Answer:

The last one

Step-by-step explanation:

Simplify and your answer is there! Like Magic! :P

Hope that helps!

kap26 [50]2 years ago

6 0

Answer:

Last one

Step-by-step explanation:

You can simplify the first one or compare both of them if they have the same ratios! Hope this helps <3

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I need help what is the sign of a. -b/b when a= 0 and b <0

kvv77 [185]

Answer:

Step-by-step explanation:

the same time as the first time since

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2 years ago

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uranmaximum [27]

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3 years ago

Please help, I will give you brainliest!

V125BC [204]

Answer:

$12 = 12 \\ 12 - 12 = 12 - 12 \\ 0 = 0$

7 0

2 years ago

Answer as many as you can please (write in slope-intercept form)

RoseWind [281]

Answer: See below

Step-by-step explanation:

For the first one, we are already given our slope. All we need to do is find the y-intercept, b.

y=-2x+b

6=-2(-3)+b

6=6+b

b=0

The slope-intercept form is y=-2x.

For the second one, we need to first find the slope using $m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$ .

$m=\frac{1-13}{3-(-6)} =\frac{-12}{9}$

Now that we have our slope, we can plug it into our slope-intercept form to solve for b.

$y=-\frac{12}{9} x+b$

$3=-\frac{12}{9}(1)+b$

$-\frac{9}{4} =b$

The slope-intercept form is $y=-\frac{12}{9} -\frac{9}{4}$ .

For the third one, we are already given the slope, so all we have to do is find b.

$y=-\frac{1}{2}x +b$

$-7=-\frac{1}{2} (-4)+b$

$-7=2+b$

$-9=b$

The slope-intercept form is $y=-\frac{1}{2} x-9$ .

For the last one, we need to first find the slope using $m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$ .

$m=\frac{-8-2}{3-1}=\frac{-10}{2} =-5$

Now that we have our slope, we can plug it into our slope-intercept form and find b.

$y=-5x+b$

$2=-5(1)+b$

$2=-5+b$

$7=b$

Our slope-intercept form is $y=-5x+7$ .

4 0

3 years ago

How do I solve: 2 sin (2x) - 2 sin x + 2√3 cos x - √3 = 0

ziro4ka [17]

Answer:

$\displaystyle x = \frac{\pi}{3} +k\, \pi$ or $\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ , where $k$ is an integer.

There are three such angles between $0$ and $2\pi$ : $\displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{2\, \pi}{3}$ , and $\displaystyle \frac{4\,\pi}{3}$ .

Step-by-step explanation:

By the double angle identity of sines:

$\sin(2\, x) = 2\, \sin x \cdot \cos x$ .

Rewrite the original equation with this identity:

$2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0$ .

Note, that $2\, (2\, \sin x \cdot \cos x)$ and $(-2\, \sin x)$ share the common factor $(2\, \sin x)$ . On the other hand, $2\sqrt{3}\, \cos x$ and $(-\sqrt{3})$ share the common factor $\sqrt[3}$ . Combine these terms pairwise using the two common factors:

$(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0$ .

Note the new common factor $(2\, \cos x - 1)$ . Therefore:

$\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0$ .

This equation holds as long as either $\left(2\, \sin x + \sqrt{3}\right)$ or $(2\, \cos x - 1)$ is zero. Let $k$ be an integer. Accordingly:

$\displaystyle \sin x = -\frac{\sqrt{3}}{2}$ , which corresponds to $\displaystyle x = -\frac{\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = -\frac{2\, \pi}{3} + 2\, k\, \pi$ .
$\displaystyle \cos x = \frac{1}{2}$ , which corresponds to $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ and $\displaystyle x = -\frac{\pi}{3} + 2\, k \, \pi$ .

Any $x$ that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:

$\displaystyle x = \frac{\pi}{3} + k \, \pi$ (from $\displaystyle x = -\frac{2\,\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ , combined,) as well as
$\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ .

7 0

3 years ago