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3 years ago

Y=2cos3x How do I graph this with labels in radians

Mathematics

1 answer:

aleksandrvk [35]3 years ago

3 0

Answer:

Linked an image

Step-by-step explanation:

Okay, basically since it is y=2co3x you know that the amplitude is 2, so rather than going to +1 and -1, you now go to +2 and -2. Since it's cosine, you also know that you start at the maximum rather than the intercept, so the starting point is at (0,2).

I'm assuming that you need to graph 2 periods, and to find the periods you divide 2pi/b, so, in this case, 2pi/3

Then lastly you need to find the four points between 0 and 2pi/3, so you divide 2pi/3/4 to get 2pi/12=pi/6

If you have any other questions just comment and I'll respond when I see it.

You might be interested in

Evaluate exactly in radians: arctan(1/sqrt(3)).

Alexeev081 [22]

$\displaystyle \\ 
 \frac{1}{ \sqrt{3} } = \frac{\sqrt{3}}{ \sqrt{3} \times \sqrt{3}} = \frac{\sqrt{3}}{ 3} \\ \\ 
\arctan\left(\frac{\sqrt{3}}{ 3} \right) = \arctan(\tan 30^o) = \boxed{30^o}$

7 0

3 years ago

Solve for x in the equation 2x^2+3x-7=x^2+5x+39

Shalnov [3]

Hey there, hope I can help!

$\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}$
$2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)$

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
$x^2-2x-46=0$

Lets use the quadratic formula now
$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$
$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}$

$\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

Multiply the numbers 2 * 1 = 2
$\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \ \left(-2\right)^2=2^2, 2^2 = 4$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \ \sqrt{4+184} \ \textgreater \ \sqrt{188} \ \textgreater \ 2 + \sqrt{188}$
$\frac{2+\sqrt{188}}{2} \ \textgreater \ Prime\;factorize\;188 \ \textgreater \ 2^2\cdot \:47 \ \textgreater \ \sqrt{2^2\cdot \:47}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \ \sqrt{47}\sqrt{2^2}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \ \sqrt{2^2}=2 \ \textgreater \ 2\sqrt{47} \ \textgreater \ \frac{2+2\sqrt{47}}{2}$

$Factor\;2+2\sqrt{47} \ \textgreater \ Rewrite\;as\;1\cdot \:2+2\sqrt{47}$
$\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \ 2\left(1+\sqrt{47}\right) \ \textgreater \ \frac{2\left(1+\sqrt{47}\right)}{2}$

$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \ 1+\sqrt{47}$

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

$\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

$\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ 2-\sqrt{188} \ \textgreater \ \frac{2-\sqrt{188}}{2}$

$\sqrt{188} = 2\sqrt{47} \ \textgreater \ \frac{2-2\sqrt{47}}{2}$

$2-2\sqrt{47} \ \textgreater \ 2\left(1-\sqrt{47}\right) \ \textgreater \ \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \ 1-\sqrt{47}$

Therefore our final solutions are
$x=1+\sqrt{47},\:x=1-\sqrt{47}$

Hope this helps!

8 0

3 years ago

Read 2 more answers

Shauna spent $175 on a pair of shoes.She spent 1/9 of the remaining money on a shirt.If he still had 4/7of his moneyleft,how muc

KIM [24]

Answer:

$490

Step-by-step explanation:

Shauna spent $175 on a pair of shoes.

She spent 1/9 of the remaining money on a shirt.

If he still had 4/7 of his money left,how much did he have at first ?

Solution:

Let at first, Shauna have = $x$

Money spent on a pair of shoes = $175

Remaining money = $x-175$

<u>As she spent 1/9 of the remaining money on a shirt.</u>

Money spent on shirt = $\frac{1}{9} \times(x-175)=\frac{x}{9} -\frac{175}{9}$

As he still had $\frac{4}{7}$ of his money left:-

Money left = $\frac{4}{7}\ of\ his\ money=\frac{4x}{7}$

Money left with Shauna = Total money, she had at first -(Money spent on a pair of shoes + Money spent on shirt )

$\frac{4x}{7} =x-(175+{\frac{x}{9} -\frac{175}{9} )\\\\\\$

$\frac{4x}{7} =x-(\frac{175}{1} -\frac{175}{9} +\frac{x}{9} )\\\\ \frac{4x}{7} =x-(\frac{1575-175}{9} +\frac{x}{9} )\\\\ \frac{4x}{7}=x-(\frac{1400}{9} +\frac{x}{9} )\\ \\ \frac{4x}{7}=x-\frac{1400}{9} -\frac{x}{9}$