What is the weight of 20 packets , if each packet weighs 3 kg.
2 answers:
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The equation relating length to width
L = 3W
The inequality stating the boundaries of the perimeter
LW <= 112
When you plug in what L equals in the first equation into the second equation, you get
3W * W <= 112
evaluate
3W^2 <= 112
3W <=
W <=
cm
L = 3W
The inequality stating the boundaries of the perimeter
LW <= 112
When you plug in what L equals in the first equation into the second equation, you get
3W * W <= 112
evaluate
3W^2 <= 112
3W <=
W <=
Quadratic Formula: (-b +/- sqrt(b^2 - 4ac)) / 2a
2x^2 - 3x = 12
2x^2 - 3x - 12 = 0
a = 2
b = -3
c = -12
(--3 +/- sqrt( (-3)^2 - 4(2)(-12) )) / 2(2)
3 +/- sqrt( 9 + 96 ) / 4
3 +/- sqrt(105) / 4
Answers: ,
Hope this helps!
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.
