What is the most common time it takes a student to say the alphabet

Mathematics

1 answer:

WITCHER [35]3 years ago

4 0

About 1 minute if you're saying it fast, and about 2 minutes if you're saying it slow.

Hope this helped☺☺

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Make graph of y=4x-7

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3 years ago

I REALLY NEED HELP WITH QUESTION!! URGENTLY!! ANYONE WHO'S GOOD AT MATHS

svet-max [94.6K]

Check the picture below.

there are a few angles shaded, I'll call that angle at O ∡O, and that angle at "x" ∡x and the green angle at A, ∡A, and the angle at B, ∡B, just so you know which angle we're referring to.

well, first off let's notice that B and C are points of tangency, meaning we get right-angles as you see in the picture, if we run an angle bisector from ∡x, towards point O, we get two congruent triangles, lemme shorten that up some, 90 + 90 + ∡O +∡x = 360, which means that ∡O + ∡x = 180, meaning that ∡O = 180 - ∡x, if that's so, the angle at point O across the shade, is 360 - ∡O or 360 - (180 - ∡x).

alrighty, by the inscribed angle theorem, ∡A is half of ∡O.

let's now focus on the triangle AOB, we'll be using half of ∡A and half of ∡O and ∡B on that triangle, let's proceed.

$\measuredangle x + \measuredangle O=180\implies \measuredangle O=180-\measuredangle x \\\\\\ \measuredangle A = \cfrac{\measuredangle O}{2}\implies \measuredangle A = \cfrac{180-\measuredangle x}{2} \\\\\\ \stackrel{\textit{the angle across from }\measuredangle O~is}{360-\measuredangle O}\implies 360-(180 - \measuredangle x)\implies 180+\measuredangle x \\\\[-0.35em] ~\dotfill$

$\stackrel{\textit{\large the sum of all three angles at }\triangle AOB}{\stackrel{\textit{half of }\measuredangle A}{\cfrac{1}{2}\cdot \cfrac{180-\measuredangle x}{2}}~~ + ~~\stackrel{\textit{half the angle across from }\measuredangle O}{\cfrac{180+\measuredangle x}{2}}~~ +~~\measuredangle B~~ =~~180} \\\\\\ \cfrac{180-\measuredangle x}{4}+\cfrac{180+\measuredangle x}{2}+\measuredangle B = 180$

now, let's multiply both sides by the LCD of all denominators, hmmmm in this case that'll be 4, so multiplying both sides by the LCD will do away with the denominators, so let's do so.

$4\left( \cfrac{180-\measuredangle x}{4}+\cfrac{180+\measuredangle x}{2}+\measuredangle B \right) = 4(180) \\\\\\ (180-\measuredangle x)+2(180+\measuredangle x)+4\measuredangle B = 720 \\\\\\ (180-\measuredangle x)+(360+2\measuredangle x)+4\measuredangle B = 720\implies 540+\measuredangle x+4\measuredangle B = 720$

$\measuredangle x+4\measuredangle B = 180\implies 4\measuredangle B = 180-\measuredangle x\implies \measuredangle B = \cfrac{180-\measuredangle x}{4} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \measuredangle B = 45 - \cfrac{\measuredangle x}{4}~\hfill$