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3 years ago

Four times a number x is at least 16

Mathematics

2 answers:

Angelina_Jolie [31]3 years ago

7 0

Answer:

Step-by-step explanation:

4x=16

x=4

egoroff_w [7]3 years ago

3 0

Answer: 4x >= 16
x >= 4

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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

Identify three pythagorean triples using the known triple 8, 15, 17. What does this mean?

ELEN [110]

Answer:

yes

Step-by-step explanation:

8² + 15² = 17²

64 + 225 = 289

289 = 289

8 0

3 years ago

Is there enough information to prove each quadrilateral is a parallelogram? Explain.

omeli [17]

Answer:

1. yes

2, no

3. yes

4. yes

Step-by-step explanation:

1. yes

If both sets of opposite sides are congruent, the quadrilateral is a parallelogram.

2. no

We know two side lengths. We know nothing about the other 2 sides and also nothing about all 4 angles.

3. yes

The missing angle must be 102°. With both pairs of opposite angles congruent, it must be a parallelogram.

4. yes

With both pairs of opposite angles congruent, it must be a parallelogram.

7 0

3 years ago

Simplify √5 • √8. A.)2√10 B.)4√10 C.)√40 D.)√13

tresset_1 [31]

Answer:

Simplify the radical by breaking the radical up into a product of known factors.

2 √ 10

Step-by-step explanation:

5 0

3 years ago

Solve the linear programming problem.

larisa [96]

Answer:

1. Objective function is a maximum at (16,0), Z = 4x+4y = 4(16) + 4(0) = 64

2. Objective function is at a maximum at (5,3), Z=3x+2y=3(5)+2(3)=21

Step-by-step explanation:

1. Maximize: P = 4x +4y

Subject to: 2x + y ≤ 20

x + 2y ≤ 16

x, y ≥ 0

Plot the constraints and the objective function Z, or P=4x+4y)

Push the objective function to the limit permitted by the feasible region to find the maximum.

Answer: Objective function is a maximum at (16,0),

Z = 4x+4y = 4(16) + 4(0) = 64

2. Maximize P = 3x + 2y

Subject to x + y ≤ 8

2x + y ≤ 13

x ≥ 0, y ≥ 0

Plot the constraints and the objective function Z, or P=3x+2y.

Push the objective function to the limit in the increase + direction permitted by the feasible region to find the maximum intersection.

Answer: Objective function is at a maximum at (5,3),

Z = 3x+2y = 3(5)+2(3) = 21

7 0

3 years ago