0.58 divided by 0.5568

867,000 rounded to the nearest hundred thousand

swat32

900,000 because it is rounded up because of how rounding works

3 0

3 years ago

What are the steps to a graph a quadratic function

Illusion [34]

You need 3 points to graph a quadratic function.

It is advisable that one of these points be the vertex, because the line of symmetry passes through it

Next, you need to find two other points that lie on the curve, replacing an x-value into the function and calculating the y-value. It is easier if these two points are at the same distance from the line of symmetry.

For example, if the quadratic function is y = (x - 4)² + 3, the vertex is located at (4, 3). Replacing with x = 3 into the equation, we get:

y = (3 - 4)² + 3

y = 1 + 3

y = 4

Replacing with x = 5 into the equation, we get:

y = (5 - 4)² + 3

y = 1 + 3

y = 4

Now, we need to connect the points (3, 4), (4, 3), and (5, 4) with a U shape, as follows:

6 0

1 year ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago

An elevator starts at the main floor and goes up 8 floors. It then goes back fown 5 floors. What integer represents elevator fin

Fed [463]

The answer is 3. He ends up on the 3rd floor.

6 0

3 years ago

Read 2 more answers

Whats the product of 6 and<br><br> q

nevsk [136]

Answer:

6q or 6 * q

Step-by-step explanation:

As you do not know what the value of the variable q is, you are essentially changing it from word form to expression form.

"product" in math means multiply, so you are multiplying 6 with q.

6q is your answer.

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7 0

3 years ago

Read 2 more answers