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Sorry, typo. \(a

Nice work  

Although I do think you should have put the first row in as well  

You can give yourself a point  

Hello Melody!

$\overline{PM} \times\overline{MR} =\overline{SM} \times \overline{MQ}$

Is there a law of geometry that says:
If a chord intersects in a circle with the diameter of the circle, the product of the sections of the chord is equal to the product of the sections of the diameter.
Does this law have a name?

I would be grateful for an explanation.
greetings

  !

Use $c = f\lambda \text{ and }E = nhf$

where c is speed of light, f is frequency, λ is wavelength, h is Planck's constant and n is number of photons.  Make sure the units are consistent.

Circle O has an area of 25pi. Line SQ is a diameter of a circle. If PM equals 6 and RM equals 3 find OM.

Hello jugoslave!

Here is the solution after the new version.

$\pi r^2=25\pi$

$r=5$

$\overline{OM}^2=3^2+5^2-2\cdot 3\cdot 5\cdot cos\ \alpha=6^2+5^2-2\cdot 6\cdot 5\cdot cos\ \alpha\\ \overline{OM}^2=34-30\cdot cos\ \alpha=61-60\cdot cos\ \alpha$

$(60-30)\cdot cos\ \alpha=61-34$

$cos\ \alpha=\frac{27}{30}=0.9$                       $\alpha = 25.8^0$

$\overline{OM}^2=34-30\cdot 0.9=34-27=7\\ \overline{OM}=\sqrt{7}$

$\overline{OM}=2.646$

  !

Ok, fair enough.

r = sqrt( 25pi / pi) = 5              PM = 6          RM = 3

I do it this way:           (r + OM) * (r - OM) = PM * RM

                                   r² - (OM)² = PM * RM

                                  25 - (OM)² = 18

                                   (OM)² = 7

                                    OM = √7   

Apology accepted (from me anyway)

However, now that you have the correct distance PM you can show us how it is done.

Have you learned from us?

I want to see not just your answer but how you have got your answer  :))

Hello, Melody! Could you please delete this post. When the question is not correct, then the answers do not make much sense. Thanks!

Melody, look at the 2nd row of your answer:

                                     (r + OM) * (r - OM) = r² - (OM)²   

I get the following:

You are right! I made a mistake. PM should be 6 NOT 7.

Correction:         Line segment PM = 6 units  

I apologize to all of you!

Well something certainly does not make sense. You are right about that. Thanks for noticing and for telling us.

 The area cannot be 25pi.     OR         PR cannot be 10

That diagram of yours is not correct. If the diameter is 10, then the cord PR cannot be 10.

Thanks asinus,

Once you know that r=5, another way to proceed is this:

$\overline{PM}*\overline{MR}=\overline{SM}*\overline{MQ}\\ 7*3=(r+\overline{OM})*(r-\overline{OM)}\\ 21=r^2-\overline{OM}^2\\ 21=25-\overline{OM}^2\\ \overline{OM}=2$

The pencils cost $0.15\ each\ and\ the\ markers\ cost\ $0.30 each. In total, Sarah spent $7.50 on pencils and markers. How many pencils did Sarah buy?

Hello Guest!

$0.15\cdot p=7.50\$-0.30\cdot m$

$p=f(m)=\frac{7.50-0.30\cdot m}{0.15}$

Sarah can buy $\large p=\frac{7.50-0.30\cdot m}{0.15}$ pencils if she has selected m markers.

  !

Circle O has an area of 25pi. Line SQ is a diameter of a circle. If PM equals 7 and RM equals 3 find OM.

Kreis O hat eine Fläche von 25 pi. Die Linie SQ ist ein Durchmesser eines Kreises. Wenn PM gleich 7 und RM gleich 3 ist, finde OM.

Hello jugoslav!

$\pi r^2=25\pi$

$r=5$

$\overline{OM}^2=3^2+5^2-2\cdot 3\cdot 5\cdot cos\ \alpha=7^2+5^2-2\cdot 7\cdot 5\cdot cos\ \alpha\\ \overline{OM}^2=34-30\cdot cos\ \alpha=74-70\cdot cos\ \alpha$

$(70-30)\cdot cos\ \alpha=74-34$

$cos\ \alpha=\frac{40}{40}=1$                       $\alpha = zero!$

$\overline{OM}^2=34-30\cdot cos\ \alpha=34-30\cdot 1=4$

$\overline{OM}=2$

  !

x + y = 82

x + 4x - 3 = 82

x = 17

y = 4x - 3 = 65

7, 1/5

A rectangular rug has a perimeter of 240 inches. The length is 12 inches more than twice the width. Find the length and width of the rug.

w + 2w + 12 = 120

w = 36

Length = 2w + 12 = 84

that doesnt seem to be right, i also got a similar answer so im not sure whats going on

Since f(x) is divisible by x^3, f(x) is of the form ax^5 + bx^4 + cx^3.

You then want ax^5 + bx^4 + cx^3 + 2 to be divisible by (x + 1)^3.  Using long division, you get the equations

-10a  + 6b - 3c = 0

4a - 3b + 2c = 0

-a + b - c + 2 = 0

==> a = 6, b = 16, c = 12

So f(x) = 6x^5 + 16x^4 + 12x^3.

1) Velocity is not speed. The mouse only went 2cm forward, that's it. His velocity is them 2cm/10 sec = 1/5 cm/s

2) Distance is 1200-150=1050 meters. So v=1050/(30*60)=7/12 m/s (Incredibly slow!)

2x+4=x+8 --> x=4. So the sides are 12,12, and 18. Drawing the altitude, by the pythagorean theorem, we have the height = sqrt(12^2-(18/2)^2)=sqrt(144-81)=sqrt(63)=3*sqrt(7). So the area is 27sqrt7