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elena-s [515]
3 years ago
11

Which parallelograms have perpendicular diagonals

Mathematics
1 answer:
REY [17]3 years ago
3 0

Answer:<em> If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. If one diagonal of a parallelogram bisects a pair of opposite angles, then the parallelogram is a rhombus. To prove a square, you must prove it is both a rectangle and a rhombus.</em>

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<h2><em /></h2>
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3x + 1 = 5X-13 if anyone knows what to do can they please help ​
Elza [17]

Answer:

  x = 7

Step-by-step explanation:

First, you look at the equation. Identify the locations of variable terms (both sides of the equal sign) and the constant terms (both sides of the equal sign).

If there are any parentheses, it is a good idea to use the distributive property to eliminate them. Here, there are none.

I like to start by subtracting <em>the variable term with the smallest coefficient</em>. Here, that is 3x, so we add -3x to both sides of the equation.

  3x -3x +1 = 5x -3x -13

  1 = 2x -13 . . . . . . . . . . . . combine like terms

Now, we have the only variable term on one side of the equal sign. We want it by itself, so we need to make the -13 go away. We do that by adding its opposite to both sides of the equation:

  1 +13 = 2x -13 +13

  14 = 2x . . . . . . . . . . . . combine like terms

Finally, we want the coefficient of 2 in the x-term to disappear. We make that happen by multiplying both sides of the equation by 1/2, the reciprocal of that coefficient.

  (1/2)(14) = (1/2)(2x)

  7 = x . . . . the solution

__

It is generally a good idea to <em>check your work</em> by seeing if your solution value satisfies the equation:

  3(7) +1 = 5(7) -13 . . . . put 7 where x is in the original equation

  21 +1 = 35 -13

  22 = 22 . . . . . . x = 7 is the solution

_____

<em>Additional comment</em>

By subtracting 3x from 5x, the result is 2x with a positive coefficient. We could solve the equation just as easily by subtracting 5x from 3x. That result would be ...

  -2x +1 = -13

Subtracting 1 would give

  -2x = -14

and you would multiply by -1/2 to get x=7. I personally like to avoid having this many minus signs show up in the problem. That is why I choose to subtract the x-term with the smallest coefficient.

5 0
4 years ago
Kent caught an 8 pound 3 ounce trout in Pike Lake. The season's record catch so far is 129 ounces. Did Kent's fish beat the reco
Crank
Yes Kent Did beat the record. There are sixteen ounces in one pound. 16•8=128+3 sooooo..131
8 0
3 years ago
Which equation matches the statement
Ulleksa [173]
-4x+2=9 would be the answer for this equation
3 0
4 years ago
Read 2 more answers
Jimmy Purchase Street blouses 3 jackets to skirts how many different outfits using a blouse a jacket and skirt a possible
Slav-nsk [51]
9 bc 3 times 3 is nine :D
6 0
3 years ago
Really need help with this pls help!!
worty [1.4K]

Let's carry this math sentence over to its natural, "shapey" element. We're going to look at each term not as an ordinary number, but as <em>the area of some shape</em>.

x² (read as "x <em>squared"</em>) can be seen as the area of a square with side lengths of x. 2x can similarly be seen as the area of a <em>rectangle </em>with a length of x and a width of 2. (Picture 1)

What's our question actually asking, though? Something about <em>perfect squares</em>. More specifically, we're looking for something to add on that'll <em>make this thing a perfect square</em>. We're trying to find a missing piece we can slot in to make a square, in other words. Problem is, our shapes don't look much like a square if we put them together right now. We need to do a little cutting and gluing first.

First, we're gonna cut the 2x rectangle lengthwise, getting two rectangles with an area of x, a length of 1, and a width of x. Next, we're going to attach them to the x² square, creating this shape that looks, strangely, like a square with a little bit missing from it (picture 2). What we're trying to do is <em>complete this square, </em>to find the area of that little missing chunk.

As it turns out, we have all the information we need for this. Notice that, using the lengths of the x rectangles, we can find that the square's dimensions are 1 x 1, which means that its area is 1 x 1  = 1.

If we tack this new area on to our original expression, we've "completed the square!" We now have a perfect square with side lengths of (x + 1) and an area of (x + 1)² (picture 3).

So, our final expression is x² + 2x + 1, and the missing constant - the area of the "missing square" we had to find to complete our larger one - is 1.

5 0
3 years ago
Read 2 more answers
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