All Answers 358087 Answers

2cos^2 theta + sin theta-2=0

$$2(1-sin^2\theta) +sin\theta-2=0\\\\
2-2sin^2\theta +sin\theta-2=0\\\\
-2sin^2\theta +sin\theta=0\\\\
sin\theta(-2sin\theta+1)=0\\\\
sin\theta=0\qquad or\qquad -2sin\theta+1=0\\\\
\theta=0,\pi,2\pi,...\;\; or\;\;sin\theta=1/2\\\\
\theta=0,\pi,2\pi,...\;\; or\;\;\pi/6,\;\; 5\pi/6\\\\
so\; for\; the \;domain\; [0,2\pi)\\\\
\theta=0,\;\frac{\pi}{6}, \;\frac{5\pi}{6}\;and\;\pi\\\\$$

you beat me Chris :(

u=-5i+3j, v=-7i-6j, w=-3i+12j 

These are not 3-dimensional vectors, so this must be just a dot product problem.

u dot w = (-5*(-3)) + (3*12) = 15 + 36 = 51.

v dot w = (-7*(-3)) + (-6*12) = 21 - 72 = -51.

51 + (-51) = 0.

EDIT: Forgot to add, thanks Phill. 

2cos^2 theta + sin theta-2=0

We can write cos^2(θ) as  1 - sin^2(θ)...so we have

2(1 - sin^2(θ)) + sin(θ) - 2 = 0

-2sin^2(θ) + sin(θ) = 0  ... multiply through by -1 and we have

2sin^2(θ) - sin(θ) = 0    factoring, we have...

sin(θ)[2sin(θ) - 1] = 0

Setting the first factor to 0, we have that sin(θ) = 0 ....... and this is true when θ = 0, pi

And setting the second factor to 0, we have 2sin(θ) - 1 = 0

Add 1 to both sides

2sin(θ) = 1   Divide by 2 on both sides

sin(θ) = 1/2   .... and this is true when θ = pi/6 and θ = 5pi/6

So our answers are  θ = [ 0, pi/6, 5pi/6, pi ] on the interval [0, 2pi)

8.50% of 15% 

0.0850 of 0.15 

0.0850 = x/0.15 

x = 0.0850*0.15

x = 0.01275.

v=3i-4j

Let's call u the unit vector, then u = v / magnitude of v

u = <3,-4> / sqrt $$(3^{2} + (-4)^{2})$$

u = <3, -4> / sqrt(25)

u = <3, -4> / 5

u = <3/5, -4/5>.

I'm not letting you have points for this Dragon!

u=6i+j and v=10i+j

u + v = (6 + 10)i + (1 + 1)j = 16i + 2j

magnitude of u+v = square root of $$(16^{2} + 2^{2})$$ 

= square root of 256 + 4 

= square root of 260

= 16.1245154965970993.

cube root of z = 8 

$$z^{3} = 8$$

$$z^{3} - 8 =0$$

$$(z-2)(z^{2}+2z+4)= 0$$

One solution is z - 2 = 0 --> z = 2.

For the other polynomial, let's use the quadratic formula where a = 1, b = 2, & c = 4: $$\begin{array}{*{20}c} {z = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{}}} \\ \end{array}$$

$$\begin{array}{*{20}c} {z = \frac{{ - 2 \pm \sqrt {4 - 16} }}{{2}}} & {{\rm{}}} \\ \end{array}$$ 

$$\begin{array}{*{20}c} {z = \frac{{-2 \pm \sqrt{-12} }}{{2}}} & {{\rm{}}} \\ \end{array}$$

We get z = -1 + 1.732050808i & -1 - 1.732050808i

The three complex roots are z = 2, z = -1 + 1.732050808i, & z = -1 - 1.732050808i.

Look here

Using the tangent inverse, we have

tan^-1 (-5/12) = -22.61986494804°......but our angle is in the second quadrant, so we must add 180° to this result...and we get...

157.38013505196°

We want to find the cosine of a first quadrant angle whose tangent = u.....  (or, u/1)

So we have....

cosΘ = 1/ √(1 + u²)

cos theta= -5/13 --> theta = cos^-1(-5/13) = acos(-5/13) = 1.965587446495.

cos(20 degrees) = 0.939692620786.

cos(pi/9) = 0.939692620786.

cos(asin(1/3) - atan(1/2)) = 0.992345241212.

The csc is just the reciprocal of the sin....so we have csc = 7/5.......

To solve this problem, we have to edit it.

$${\frac{{\frac{{\mathtt{61}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{31}}}{{\mathtt{3}}}}$$

Now, we multiply 61/2 by 31.

$${\frac{{\mathtt{61}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{31}} = {\mathtt{945.5}}$$

The answer can be $${\mathtt{945.5}}$$, or 945 1/2.

But, let's just keep $${\mathtt{945.5}}$$ for now.

Now, divide 945.5 by 3.

$${\frac{{\mathtt{945.5}}}{{\mathtt{3}}}} = {\mathtt{315.166\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

Because the number is too big, let's round it.

$${\mathtt{315.166\: \!666\: \!666\: \!666\: \!666\: \!7}} = {\mathtt{315.2}}$$

So, when we round our answer, it is $${\mathtt{315.2}}$$.

Hope it helps!

To find the answer, group them first.

349,615

Now, let's round. We are rounding to the nearest thousand, so the thousand is 9,000.

350,000

349,615

349,000

When we round REMEMBER that if a number below 5 is the same, and that if a number is 5 and above, it goes up on.

So, the answer is $${\mathtt{350\,000}}$$. Hope it helps!

10p²-15p-25 

I assume you might want to factor this??

First, take out the greatest common factor, 5.....so we have

5(2p² - 3p -5)   and factoring the expression in the parentheses, we have

5(2p - 5)(p + 1)

Now if you want to find the roots of this, we simply set each factor equal to 0 and we have that p = 5/2 or p = -1

Thanks, AzizHusain.

Yes, it will arrive over the branch so the bear cannot reach the food. You can tie it as well.

I am imagining something like this:

$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{60}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}$$

You have to use the order of operations to do this problem.

An easy way to remember it is PEMDAS, or PEDMAS.

PEDMAS = Parentheses, Exponents, Division, Multiplication, Addition, & Subtraction.

PEMDAS = Parentheses, Exponents, Multiplication, Division, Addition, & Subtraction. 

Now let's multiply the question one by one.

$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{60}} = {\mathtt{180}}$$

$${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{9}} = {\mathtt{45}}$$

$${\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}} = {\mathtt{4}}$$

Now that we got everything, let's add it.

$${\mathtt{180}}{\mathtt{\,\small\textbf+\,}}{\mathtt{45}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}} = {\mathtt{225}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}} = {\mathtt{229}}$$

So, our final answer is $${\mathtt{229}}$$.

It is simple as that.

And on a line graph?

(That's what I meant)

But Thanks, AzizHusain.

To find the value of n, we have to substitute n into 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11...

Now let's substitute n.

$${\mathtt{2}}{n} = {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{3}}$$

In this situation, n = 3.

So, let's solve the problem.

$${\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{3}} = {\mathtt{6}}$$

So, $${\mathtt{6}}$$ is our answer.

Thank you, AzizHusain. I needed to know that. I was told in the math book to do a line graph, but I didn't post it here.

Can you please do the line graph for me please?

Thanks.