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Sati [7]
3 years ago
13

How do you do 5÷465 in long division and 12÷924 and 3÷1107

Mathematics
1 answer:
horsena [70]3 years ago
6 0

Answer:1st one 93 2nd 77 and 3rd 369

Step-by-step explanation:

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The radius of circle L is 22 cm.
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2R = 1D
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4 years ago
No idea what this means​
GenaCL600 [577]

Answer:

basically just find the hypotenuse and show your setup/expression

a²+b²=c²

Step-by-step explanation:

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3 years ago
Model and solve the following<br>equations:<br>3x +3 = 2x + 7<br>5x + 2 = 3x = 12<br>​
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hope it helps you!!!!!!!!!!

4 0
3 years ago
CAN SOMEONE GOOD AT MATH HELP PLEEESE!!
Rufina [12.5K]

Answer: 1 .Thus for a graph to have an Euler circuit, all vertices must have even degree. The converse is also true: if all the vertices of a graph have even degree, then the graph has an Euler circuit, and if there are exactly two vertices with odd degree, the graph has an Euler path.

2. A graph has an Euler circuit if and only if the degree of every vertex is even. A graph has an Euler path if and only if there are at most two vertices with odd degree.

Step-by-step explanation:

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7 0
3 years ago
I don't know if this is right... please someone help mee
worty [1.4K]
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.
6 0
3 years ago
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