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kotegsom [21]
2 years ago
13

A multiple of 3 but not of 6 and 9, greater than 22 but less than 37​

Mathematics
2 answers:
DENIUS [597]2 years ago
8 0

Answer:

33

Step-by-step explanation:

33 is not a multiple of 6 because 36 is.

33 is not a multiple of 9 because 36 is also.

V125BC [204]2 years ago
7 0

Answer:

i think it's 33

Step-by-step explanation:

33 is a multiple of 3, not a multiple of 6 or 9

33/6 = 5.5

33/9 = 3.666666666667

33/3 = 11

and it is less than greater than 22 and less than 37

hope this helped!

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For the first circle, let's use the pythagorean theorem

\bf \textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;c=\sqrt{8^2+15^2}\implies c=\sqrt{289}\implies c=17

now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.

 so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.

when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.



now, let's check for second circle

\bf \textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;c=\sqrt{11^2+14^2}\implies c=\sqrt{317}\implies c\approx 17.8044938

well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.



let's check the third circle

\bf \textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;c=\sqrt{33^2+56^2}\implies c=\sqrt{4225}\implies c=\stackrel{33+32}{65}

this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.
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3 years ago
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