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Anni [7]
3 years ago
13

Which of the following equations have complex roots?

Mathematics
1 answer:
valkas [14]3 years ago
8 0

Answer:

B

Step-by-step explanation:

A:

Completing the square:  (x+ 7/2)^2 = -2 + 49/4 = 41/4, in the next step you'll be square rooting a positive number so it'll have real roots.

B:

Completing the square:  (x+3/2)^2 = -9 + 9/4 = -27/4, you'll next be square rooting a negative number, so it'll have complex roots

C:

Completing the square:  (x-5/2)^2 = -1 + 25/4 = 21/4, next you'll square root a positive number, so real roots

D:

Completing the square: (x+7/2)^2 = 2 + 49/4 = 57/4, square root a positive number, so real roots.

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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
Above is a table with missing terms. Come up with a possible common difference that makes sense and list the missing terms.
Rina8888 [55]

Answer:

common ratio:2 2. 18  3.36  4.72 equation: 4.5(2^x)

Step-by-step explanation:

144=2*2*2*2*3*3    9=3*3

the common ratio would be 2

2. 9*2=18

3. 18*2=36

4. 36*2=72

(check) 72*2=144

0. 9/2=4.5

equation: 4.5(2^x)

7 0
2 years ago
2. Suppose the temperature that most foods can stay bacteria free in restaurants varies approximately according to a normal dist
vagabundo [1.1K]

Answer:

28.4

Step-by-step explanation:

Given that:

Mean, m = 31.3

Standard deviation, s = 2.8

Since, data is normally distributed :

P(x < 0.15) gives a Z value of - 1.036

Using the Zscore formula :

Z = (x - mean) / standard deviation

-1.036 = (x - 31.3) / 2.8

-1.036 * 2.8 = x - 31.3

-2.9008 = x - 31.3

-2.9008 + 31.3 = x

28.3992 = x

The temperature which correlates to the bottom 15% of the distribution is 28.4

7 0
2 years ago
A pencil is `120` millimeters long.
leva [86]

Given the length of the pencil, the length of the marker is 130.8 millimeters.

<h3>What are percentages?</h3>

Percentage can be described as a fraction of an amount expressed as a number out of hundred. Percentages are represented %.

<h3>How long is the marker?</h3>

Length of the marker = (1 + percentage) x length of the pencil

(1.09) x 120 = 130.8 millimeters

To learn more about percentages, please check: brainly.com/question/25764815

6 0
2 years ago
name the set of 6 consecutive integers starting with -3 put the set in braces and put commas between the elements of the set
melisa1 [442]
6-3=3 that is the answer ok
6 0
3 years ago
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