Why can't you use tangent and secant theorem

*Why didn't you use decimals?*

Your video title and description needs rewording. It is a mess.

I am so pleased😢🎉❤😂

Nice problem ❤

Dear, your email please I wish to congratulate you for this construction. Anyway CONGRATULATIONS!

Yes. Draw two vertical lines from the circle centers to the 18 cm side. The distance between them is 18 - 5 - 8 = 5 cm. Form a right triangle with hypotenuse between circle centers. The height of the triangle is Sqrt( ((8 + 5) ^2) - 5^2) = sqrt(169 - 25) = sqrt(144) = 12. Then X = 12 + 5 + 8 = 25 and the Area of the rectangle is 18 * 25 = 20 * 25 - 50 = 450 cm ^2. I'll watch the video and see if we agree.

Because lines OS and TQ are parallel, you can add them to get 7. OQ would be the hypotenuse. (7)^2 + b^2 = (14)^2. b=√149 =7√3.

For

What was the point of the circles?

Alternative method: ∠PYX = 180° - (56° + 80°) = 44° By the alternate segment theorem ∠SPX = 44° ∠SPX and ∠PQY are corresponding angles ∴ ∠PQY = 44°

35

All are OK. Go on.

Good approach.

*It's easy to show that this diagram is impossible. Since angle QRS (which is the same as angle PRS) is 40°, that means that angle QOS is 80° due to the inscribed angle theorem. The radius OS is perpendicular to the tangent PS, so the angle PSO would be 90°. But then the triangle PSO would have angles of 40°, 80° and 90°, which is impossible as the sum of the interior angles of a triangle **_must_** be 180° and 40+80+90=210°.*

Neither solution shows where TQP=35 comes from.

∠QRP = 35° so by the inscribed angle theorem ∠QOP = 70° TQ is a tangent to the circle, so TQ⊥QO ∴ △TQO is right-angled with ∠TQO = 90° so by the properties of a △: ∠QTO =180 - (90 + 70) = 20°

Ang PQR+angPSR=180 Ang PSR+81+53=180 Hence Any PQR +ang PSR=ang PSR + 81+53 Rejecting PSR from both sides Ang PQR=81+53=134 degrees

Hi! Suggestion for a slightly more satisfying solution because it involves smaller numbers and immediate mental resolution (very subjective criteria, I know): QPR = 55 and TQP = 35, as you noted QPR is an external angle -> QPR = PTQ + PQT -> 55 = x+35

Inverse of this function shouldn't exist, it is not a one-one function

Why ±

Thanks a lot professor I follow you from

Easier way: 4 quarter circles = 1 whole circle. The radius r=2/2=1, so now C=2πr=2π(1)=2π cm

Before I watch the video, here's my work: Would not the perimeter be the same as the circumference of a full circle with radius 1 cm since it is equal to the sum of the arc lengths of 4 equally-sized quarter circles of radius 1cm? The circumference of a circle of radius 1 cm=2pi cm. There, final answer. Now to see if my work is correct... Edit for post-video thoughts: He did the math correctly. However, I think he missed a simpler rational in his explanation (that is based in some of the properties of the figure he mentioned. Namely the rational that since each of the four arc lengths is that of a quarter circle with equal radii, the sum of the arc lengths would be that of the circumference of a full circle of equal radius as I said above.

Okay, so we have 5*5^(2x)+5^(2x)=150 150/6*=25. 2x=2, so x=1. *If anyone's wondering how I got 6*5^(2x)=150, I simply used the property that ax+x=(a+1)x..

1

Bakıdan salamlar.Əla həll etdiniz.

70

Been a while since i learned inverse functions. I needed this recap hehe

❤

Thanks sir

Thanks for sharing you knowledge. Your way of solving it was very smart. I liked a lot!

That is a lot easier than what I was trying to do :D

Draw SQ. SPR = SQR = 81 (Angles subtended by chord are equal) PRS = PQS = 53 (as above) PQR = PQS + SQR = 81 + 53 = 134

Is it 40 sir

134 easy

Well solution sir

This is literally what teacher told us how to do these fractions. a/b+c/d=(a*d+c*b)/b*d.

Massive

Wow this is brilliant👏👏

HOW TO FIND THE EXACT EXACT EXACT CIRCUMFERENCE OF CIRCLE . Procedure Step 1. Draw a square, 2 diagonals and inscribe a circle in the square. Thereby side and diameter will be the same Step 2 : Subtract 2 diagonals from the perimeter of square Step3. Divide step2 with 8 Step4. Add 3 times of the side. to Step 3 At the end we get the EXACT circumference of the inscribed circle

= 360 degree in front of the value of rs or the text of pm January work sheet 32 22 y 49 days ago

Nice trick

So interesting

I paused the video just before the end, and arrived at 36 V. This is very good information, since the gate-source capacitance of MOSFET's are rated in uC. They usually rate them at 10v. which makes the calculation quite easy. It's much easier to determine gate drive impedance, being able to convert to capacitance.

Nice lesson

Nice maths trick but can you show us a trick using much bigger numbers

/: if there is a zombie apocalypse she would be the only one not gone for

Great

Laws of Logarithms