The perimeter of a rectangular field is 232 yards. if the length of the field is 65 yards, what is its width?

Which ordered pair represents the plotted point on the graph?

mezya [45]

Hope this helps but the answer is A.(-5,2)

7 0

2 years ago

Use the law of syllogism to form a conclusion from the given premises. Premise 1: If a polygon was translated to the right, then

Pachacha [2.7K]

<span>Premise 1: If a polygon was translated to the right, then its image is congruent to its pre-image.

In symbols: p => q

Premise 2: If an image is congruent to its pre-image, then a rigid transformation was performed.

In symbols: q => r

By the law of silogism

(p => q) and (q => r) => q => r

So, the conclusion is that </span><span>if a polygon was translated to the right, then a rigid transformation was performed. <---- answer</span>

3 0

2 years ago

Read 2 more answers

What is the value of m in the equation 5m − 7 = 6m 11? 18 1 −18 −1.

riadik2000 [5.3K]

Answer: 18

Step-by-step explanation: 5m - 7 = 6m + 11

Solve for M

5m - 6m = 11+ 7

-m = 18

Divide both sides by -1

-m/-1 = 18/-1

m = -18

4 0

2 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

A cone and a cylinder have equal radius, r, and equal altitudes, h. If the slant height of the cone is l, then the ratio of the

77julia77 [94]

Given:

radius of cone = r

height of cone = h

radius of cylinder = r

height of cylinder = h

slant height of cone = l

Solution

The lateral area (A) of a cone can be found using the formula:

$A\text{ = }\pi rl$

where r is the radius and l is the slant height

The lateral area (A) of a cylinder can be found using the formula:

$A\text{ = 2}\pi rh$

The ratio of the lateral area of the cone to the lateral area of the cylinder is:

$\pi rl\text{ : 2}\pi rh$

Canceling out, we have:

$\begin{gathered} =\frac{\pi rl}{2\pi rh} \\ =\text{ }\frac{l}{2h}\text{ or l:2h} \end{gathered}$

Hence the Answer is option B

3 0

10 months ago