The answer is 6 units and 3 fourths
Answer:
15x^2 - 12x^3
Step-by-step explanation:
A rectangular block has 3 parts that play into its volume. length, width and height. The question gives us length and width in the form of x and 3x, so height is what's missing.
It gives us a bit more information saying the sum of its edges is 20. We also have to ask how many lengths, widths and heights are there. That may be a bit hard to understand, but is you are looking at a block I could ask how many edges are vertical, just going up and down. These would be the heights. There are 4 total, and this goes the same for length and width, so 4*length + 4*width and 4*height = 20.
Taking that and plugging in x for length and 3x for width (or you could do it the other way around, it doesn't matter, you get:
4*x + 4*3x + 4*height = 20
4x + 12x + 4h = 20
16x + 4h = 20
4h = 20 - 16x
h = 5 - 4x
Now we have h in terms of x, which lets us easily find the volume just knowing x. To find the volume of a rectangular block you just multiply the length, width and height.
x*3x*(5-4x)
3x^2(5-4x)
15x^2 - 12x^3
Question doesn't give a specific value for x at all so you should be done there. Any number you plug in for x should get you the right answer
Answer:
A) I only
Step-by-step explanation:
median = 25
mean = 36
Plot for given distribution is shown in fig attached below. mean is shown with red block and median with green block. plot is skewed to the left and there is no outlier.
Answer:
As you noted, ∠PRS=80°. Take the triangle △TRS: we know two angles out of three, so that ∠TSR=45°. Now take the isosceles triangle △PTS: the two angles adjacent to the base PT are equal to 35°, so the third angle ∠PST=110° and then ∠PSR=65°. This is one of the two angles adjacent to the base of an isosceles trapezoid: knowing it, you can easily complete the solution.
Hi!
Yes, it is. When we have one inequality composed of two inequalities, we call this a double inequality.
A double inequality would look something like <em>-5 < 4x < 10.
</em>Hopefully, this helps! =)<em>
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