just like the previous one, we sum the sides to get the perimeter.
Use the side lengths! C and D have lengths of 3 and heights of 4.
Also use how far away they are from the axis.
Both C and D are 2 away from each axis on the end point.
This also goes with A.
However, B does not seem to be congruent in the sense of (coordinates)
It is the same length and height.
Answer:
I think its A i'm sorry if its wrong
Step-by-step explanation:
Answers:
- Discrete
- Continuous
- Discrete
- Continuous
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Explanations:
- This is discrete because we can't have half a basketball, or any non-whole decimal value to represent the number of basketballs. We can only consider positive whole numbers {1,2,3,4,...}. A discrete set like this has gaps between items. In other words, the midpoint of 2 and 3 (the value 2.5) isn't a valid number of basketballs.
- This is continuous because time values are continuous. We can take any two different markers in time, and find a midpoint between them. For example, the midpoint of 5 minutes and 17 minutes is 11 minutes since (5+17)/2 = 22/2 = 11. Continuous sets like this do not have any gaps between items. We can consider this to be densely packed.
- This is the same as problem 1, so we have another discrete function. You either score a bullseye or you don't. We can't score half a bullseye. The only possible values are {1,2,3,4,...}
- This is similar to problem 2. This function is continuous. Pick any two different positive real numbers to represent the amount of gallons of water. You will always be able to find a midpoint between those values (eg: we can have half a gallon) and such a measurement makes sense.
So in short, always try to ask the question: Can I pick two different values, compute the midpoint, and have that midpoint make sense? If so, then you're dealing with a continuous variable. Otherwise, the data is discrete.
Answer:
Step-by-step explanation:
Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).
Let's calculate the common points:
![x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1](https://tex.z-dn.net/?f=x%5E2%3D%5Csqrt%7Bx%7D%5Cqquad%5Ctext%7Bsquare%20of%20both%20sides%7D%5C%5C%5C%5C%28x%5E2%29%5E2%3D%5Cleft%28%5Csqrt%7Bx%7D%5Cright%29%5E2%5C%5C%5C%5Cx%5E4%3Dx%5Cqquad%5Ctext%7Bsubtract%7D%5C%20x%5C%20%5Ctext%7Bfrom%20both%20sides%7D%5C%5C%5C%5Cx%5E4-x%3D0%5Cqquad%5Ctext%7Bdistribute%7D%5C%5C%5C%5Cx%28x%5E3-1%29%3D0%5Ciff%20x%3D0%5C%20%5Cvee%5C%20x%5E3-1%3D0%5C%5C%5C%5Cx%5E3-1%3D0%5Cqquad%5Ctext%7Badd%201%20to%20both%20sides%7D%5C%5C%5C%5Cx%5E3%3D1%5Cto%20x%3D%5Csqrt%5B3%5D1%5Cto%20x%3D1)
The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.
We have integrals:
![\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}](https://tex.z-dn.net/?f=%5Cint%5Climits_%7B0%7D%5E1%28%5Csqrt%7Bx%7D%29dx-%5Cint%5Climits_%7B0%7D%5E1%28x%5E2%29dx%3D%28%2A%29%5C%5C%5C%5C%5Cint%28%5Csqrt%7Bx%7D%29dx%3D%5Cint%5Cleft%28x%5E%5Cfrac%7B1%7D%7B2%7D%5Cright%29dx%3D%5Cdfrac%7B2%7D%7B3%7Dx%5E%5Cfrac%7B3%7D%7B2%7D%3D%5Cdfrac%7B2x%5Csqrt%7Bx%7D%7D%7B3%7D%5C%5C%5C%5C%5Cint%28x%5E2%29dx%3D%5Cdfrac%7B1%7D%7B3%7Dx%5E3%5C%5C%5C%5C%28%2A%29%3D%5Cleft%28%5Cdfrac%7B2x%5Csqrt%7Bx%7D%7D%7B2%7D%5Cright%5D%5E1_0-%5Cleft%28%5Cdfrac%7B1%7D%7B3%7Dx%5E3%5Cright%5D%5E1_0%3D%5Cdfrac%7B2%281%29%5Csqrt%7B1%7D%7D%7B2%7D-%5Cdfrac%7B2%280%29%5Csqrt%7B0%7D%7D%7B2%7D-%5Cleft%28%5Cdfrac%7B1%7D%7B3%7D%281%29%5E3-%5Cdfrac%7B1%7D%7B3%7D%280%29%5E3%5Cright%29%5C%5C%5C%5C%3D%5Cdfrac%7B2%281%29%281%29%7D%7B2%7D-%5Cdfrac%7B2%280%29%280%29%7D%7B2%7D-%5Cdfrac%7B1%7D%7B3%7D%281%29%7D%2B%5Cdfrac%7B1%7D%7B3%7D%280%29%3D2-0-%5Cdfrac%7B1%7D%7B3%7D%2B0%3D1%5Cdfrac%7B1%7D%7B3%7D)
