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Paha777 [63]
3 years ago
5

Which method is the best method for solving the equation?

%20%3D%200" id="TexFormula1" title="8x^{2} - 13x + 3 = 0" alt="8x^{2} - 13x + 3 = 0" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Paladinen [302]3 years ago
3 0

Answer:

Step-by-step explanation:

Which Method Should Be Used to Solve a Quadratic Equation? Quadratic equations are of the form ax2 + bx + c = 0 where a, b and c are real numbers, a ≠ 0. Quadratic equations have two solutions, but it is possible that one solution may repeat. happy to help

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Step-by-step explanation:

If he read 9 books in january and read 4 more in july than he did in january then he read 13 books in july.

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HELP PLEASE ILL VENMO THE FIRST PERSON WHO ANSWERS
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Answer:

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Step-by-step explanation:

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3 years ago
Please Solve This For Me If You Can y(2.5+7)+y-2 ​
Law Incorporation [45]

Answer: 10.5y-2

Hope this helps!

4 0
3 years ago
Read 2 more answers
How do you do this question?
Ksivusya [100]

Answer:

V = (About) 22.2, Graph = First graph/Graph in the attachment

Step-by-step explanation:

Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.

\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}

The plus 1 in '1 + 2/x' is shifting this graph up from where it is rotating, but the negative 1 is subtracting the area between the y-axis and the shaded region, so that when it's flipped around, it becomes a washer.

V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\

\int _1^3\left(1+\frac{2}{y}\right)^2dy=4\ln \left(3\right)+\frac{14}{3}, \int _1^31dy=2\\\\=> \pi \left(4\ln \left(3\right)+\frac{14}{3}-2\right)\\=> \pi \left(4\ln \left(3\right)+\frac{8}{3}\right)

Our exact solution will be V = π(4In(3) + 8/3). In decimal form it will be about 22.2 however. Try both solution if you like, but it would be better to use 22.2. Your graph will just be a plot under the curve y = 2/x, the first graph.

5 0
3 years ago
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