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3 years ago

What value is added to both sides of the equation x2 − 6x = 10 in order to solve by completing the square?

1 answer:

8 0

$x^2-6x=10\\\\x^2-2\cdot x\cdot3=10\ \ \ |+3^2=9\\\\\underbrace{x^2-2\cdot x\cdot3+3^2}_{(a-b)^2=a^2-2ab+b^2}=19\\\\(x-3)^2=19$

Answer: 9

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Answer:

B. 3 people;9 cups

Step-by-step explanation:

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2 years ago

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1.4t−0.4(t−3.1)=5.8 SOLVE FOR Y

VladimirAG [237]

There is no y in this question, however we can solve for t.

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3 years ago

Kim is reading a book with 380 pages. She read 20 pages each day until she reached Part 2 of the book. Part 2 of the book is 160

o-na [289]

Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

The base of a solid is the region in the first quadrant between the graph of y=x2 and the x -axis for 0≤x≤1 . For the solid, eac

nata0808 [166]

Answer:

The volume of the solid is π/40 cubic units.

Step-by-step explanation:

Please refer to the graph below.

Recall that the area of a semi-circle is given by:

$\displaystyle A=\frac{1}{2}\pi r^2$

The volume of the solid will be the integral from x = 0 to x = 1 of area A. Since the diameter is given by y, then the radius is y/2. Hence, the volume of the solid is:

$\displaystyle V=\int_0^1\frac{1}{2}\pi \left(\frac{y}{2}\right)^2\, dx$

Substitute:

$\displaystyle V=\frac{1}{2}\pi\int_0^1\left(\frac{x^2}{2}\right)^2\, dx$

Simplify:

$\displaystyle V=\frac{1}{2}\pi \int_0^1\frac{x^4}{4}\, dx$

Integrate:

$\displaystyle V=\frac{1}{2}\pi \left[\frac{x^5}{20}\Big|_0^1\right]$

Evaluate:

$\displaystyle V=\frac{\pi}{40}\left((1)^5-\left(0\right)^5\right)=\frac{\pi}{40}\text{ units}^3$

The volume of the solid is π/40 cubic units.

4 0

3 years ago