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Butoxors [25]
4 years ago
10

Simplify this expression. 6x2(3x) A) 18x2 B) 18x3 C) 108x2 D) 108x3

Mathematics
2 answers:
Anon25 [30]4 years ago
8 0
6x2 = 12
12x3 = 36
Answer:
A) 18x2 = 36
juin [17]4 years ago
5 0
= 6x^2 * 3x
= 18x^3

its best to write 6x2  as 6x^2  to avoid confusion
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Bad White [126]
5 1/4 - 3 3/4 = 4 5/4 - 3 3/4 = 1 2/4
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4 years ago
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The ratio of the sides of 2 cubes is 2 to 7. If the volume of the smaller cube is 32 u3, then the volume of the larger cube is _
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\bf \qquad \qquad \textit{ratio relations}
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\begin{array}{ccccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array} \\\\
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\cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
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\bf \cfrac{small}{large}\qquad \cfrac{s}{s}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\implies \cfrac{2}{7}=\cfrac{\sqrt[3]{32}}{\sqrt[3]{v}}\implies \cfrac{2}{7}=\sqrt[3]{\cfrac{32}{v}}\implies \left( \cfrac{2}{7} \right)^3=\cfrac{32}{v}
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\cfrac{2^3}{7^3}=\cfrac{32}{v}\implies v=\cfrac{7^3\cdot 32}{2^3}
5 0
3 years ago
Usian bolt can run 100meters in 9.58 seconds . my car can drive 100kilometers an hour. which is faster, is it faster to travel 1
Mashcka [7]

Answer:

Your car is faster of course

Step-by-step explanation:

100 km would be 100000 meters. Then you would need to know that if Usain bolt ran 100 meters every 9.54 sec then in 100000 meters he would have ran 9540 seconds which is also known has 159 minutes, which is more than 1 hour so your car is faster.

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3 years ago
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100. This is very easy thanks teacher
6 0
2 years ago
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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.
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3 years ago
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