A paint roller, shown below, is used for painting large surfaces.

Which statement best explains how Lizabeth develops the theme in "Marigolds" that beauty can be appreciated even in the worst ci

Anika [276]

That no matter what if you feel good about yourself you do better at what your doing. Hope I helped!

4 0

2 years ago

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4 raised to power two minus 8u minus 9=0

coldgirl [10]

Answer:

x can be 3, -3, i, or -i.

Step-by-step explanation:

If you can't find the factoring by looking at this, simplify the equation.

let

x

2

=

y

to make things easier to see

Now we have

y

2

−

8

y

−

9

=

0

See it now?

We can factor into

(

y

−

9

)

(

y

+

1

)

Now substitute back in

x

2

for

y

.

(

x

2

−

9

)

(

x

2

+

1

)

Since

(

x

2

−

9

)

is a difference of two squares,

(

x

−

3

)

(

x

+

3

)

Now we have

(

x

−

3

)

(

x

+

3

)

(

x

2

+

1

)

x can be 3, -3 for the first two parts

x

2

+

1

=

0

can become

x

2

=

−

1

Taking the positive and negative root means

x

=

±

√

−

1

Thus

x

=

±

i

in addition to 3 and -3.

4 0

3 years ago

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The sides of a triangle measures 6 inches, 8inches, and 10inches. The triangle has one 90 degree angle. What type of triangle is

IrinaVladis [17]

It's a right triangle as it has the 90 degree corner and 3 different lengths

5 0

2 years ago

What do you mean by this symbol ^

12345 [234]

Answer:

it means the power of

Step-by-step explanation:

like 2^2= 4

3^2= 9

3 0

2 years ago

Read 2 more answers

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

or

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

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

5 0

3 years ago