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swat32
3 years ago
15

Jeremy drew a polygon with four right angles and four sides with the same length.what kind of polygon did jeremy draw?

Mathematics
1 answer:
Naily [24]3 years ago
8 0
This is very easy! It's a square.
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In order to avoid heavy traffic, Antonio can drive home after working less than 7 hours (h < 7) or after working more than 9
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The Answer is B = 7 > h > 9...
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How far is it from A to C?<br> D<br> 25 yd<br> 40 yd<br> A<br> E<br> F<br> 60 yd<br> x yd
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2 years ago
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2 years ago
Two tourists went on a hike at dawn. One went from a to b and another one went from b to a. They met at noon but did not stop an
Alecsey [184]
<span>Dawn was at 6 am. Variables a = distance from a to passing point b = distance from b to passing point c = speed of hiker 1 d = speed of hiker 2 x = number of hours prior to noon when dawn is The first hiker travels for x hours to cover distance a, and the 2nd hiker then takes 9 hours to cover that same distance. This can be expressed as a = cx = 9d cx = 9d x = 9d/c The second hiker travels for x hours to cover distance b, and the 1st hiker then takes 4 hours to cover than same distance. Expressed as b = dx = 4c dx = 4c x = 4c/d We now have two expressions for x, set them equal to each other. 9d/c = 4c/d Multiply both sides by d 9d^2/c = 4c Divide both sides by c 9d^2/c^2 = 4 Interesting... Both sides are exact squares. Take the square root of both sides 3d/c = 2 d/c = 2/3 We now know the ratio of the speeds of the two hikers. Let's see what X is now. x = 9d/c = 9*2/3 = 18/3 = 6 x = 4c/d = 4*3/2 = 12/2 = 6 Both expressions for x, claim x to be 6 hours. And 6 hours prior to noon is 6am. We don't know the actual speeds of the two hikers, nor how far they actually walked. But we do know their relative speeds. And that's enough to figure out when dawn was.</span>
8 0
3 years ago
When x-y=3 work out the value of y-x
GalinKa [24]

The value of y-x is (-3)

<u>Solution:</u>

Given: The value of x-y=3

To find: The value of y-x

\bold{x-y=3\rightarrow (1)}

Let's multiply the equation (1) by (- 1) on both sides,

\Rightarrow\bold{(x - y)\times(- 1) = 3\times(- 1)}

\Rightarrow\bold{-(x-y)=-3}

On multiplying the sign,

\Rightarrow\bold{(-x+y)=-3}

The above equation can also be written as,

\Rightarrow\bold{y-x=-3\rightarrow \text{(The required equation)}}

<u>Multiplication of signs:</u>

\Rightarrow( + )\times( + ) = ( + )

\Rightarrow( + )\times( - ) = ( - )

\Rightarrow( - )\times( + ) = ( - )

\Rightarrow( - )\times( - ) = ( + )

In simpler terms, when we multiply two integers with the same signs, the result is always positive and when we multiply two integers with different signs, the result is always negative.

5 0
3 years ago
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