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3 years ago

Which statement explains how you could use coordinate geometry to prove that quadrilateral ABCD is a square

Mathematics

1 answer:

mixer [17]3 years ago

3 0

Answer:

Prove that all sides are congruent and that the slopes of consecutive sides are opposite reciprocals. Step-by-step explanation: In order to two segmets to be perpendicular, the slope of both lines must be opposite and reciprocals, having a 90° interception and forming a square.

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Work out the Perimeter of the Triangle For 100 Points

WINSTONCH [101]

Answer:

P = 4x + 14

Step-by-step explanation:

$P=(3x-5)+(19-x)+(2x)=(3x-x+2x)+(-5+19)=4x+14$

Hope this helps

5 0

2 years ago

Thanks for any help provided to my question! <3<br>5m + 2(m + 8) + 3

disa [49]

5m+2(m+8)+3
5m+2m+16+3
7m+19

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3 years ago

The point (-2, -1) satisfies which of the following inequalities?

satela [25.4K]

Answer:

Step-by-step explanation:

4 0

3 years 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

Is x = 7 a solution to the equation 3x - 4 = 15? No Yes

Elis [28]

Answer:

No, because 3 x 7 = 21 - 4 = 17

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers