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vladimir1956 [14]

3 years ago

15

Mitch can grow 150 wild flowers with every seed packet. Write an equation that shows the relationship between the number of seed

packets (p) and the total number of wild flowers (w).

1 answer:

Troyanec [42]3 years ago

7 0

P = 150w

In words, 1 seed packet (p) is 150 wild flowers (w)

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I’ll make you brainlist thankss

dusya [7]

Answer:

The equation of the line will be:

y = 0

Step-by-step explanation:

Given the points

(0, 0)

(3, 0)

Finding the slope

$\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(0,\:0\right),\:\left(x_2,\:y_2\right)=\left(3,\:0\right)$

$m=\frac{0-0}{3-0}$

$m=0$

We know that the value of y-intercept can be obtained by setting x=0, and solving for y

Here,

at x = 0, the value of y = 0

Thus,

y-intercept = b = 0

We know the slope-intercept form of the line equation is

$y=mx+b$

where m is the slope and b is the y-intercept

now substituting m = 0 and b = 0 in the slope-intercept form

y = mx+b

y = 0(x) + (0)

y = 0+0

y = 0

Therefore, the equation of line will be:

y = 0

8 0

3 years ago

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:

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

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

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

Find the equation of the line specified.

OLga [1]

The point slope form of a straight line is y-y1 = m(x-x1)

y -(-3) = -7(x-5)

y + 3 = -7(x-5)

y = -7x + 35 - 3

y = -7x + 32

7 0

3 years ago

Insert three arthimetic means between 4 and 10

lapo4ka [179]

Yes I work this out so yea you divide idk

7 0

3 years ago

Read 2 more answers

Tobias has 12 coins in his pocket. Of these coins, 8 were made in the year 2000,and

Free_Kalibri [48]

Answer on attached file.
Bye.

7 0

4 years ago