All categories

3 years ago

A bicyclist travels 1 miles in 5 minutes .If m represents minutes , what does the expression m/5 represents

Mathematics

1 answer:

notsponge [240]3 years ago

5 0

It represents miles per minute or in the case 1/5, or m/5

You might be interested in

Joshua took a total of 12 "i-Ready" quizzes over the course of 2 weeks. How many weeks of school will Joshua have to attend this

Sergio039 [100]

I think the answer is 2 weeks 6 days because if over the course of 2 weeks they took 12 I ready quizzes the he needs 18 quizzes and it leaves 1 day extra

7 0

3 years ago

82 decrease by 10 percent

zzz [600]

Answer:

Percentage decrease = 10% × 0.82

New value = 0.82 - Percentage decrease

Let me know if this helps! :)

8 0

2 years ago

Evaluate log base 16 0.25

AveGali [126]

Answer:

Step-by-step explanation:

Rewrite log16(0.25)

using the change of base formula.

Exact Form:

log(0.25)log(16)

Decimal Form:

−0.5

5 0

3 years ago

Identify the type of function represented by f(x)=3(1.5)^x

Murljashka [212]

Here your base (1.5) is greater than zero, so you have exponential growth here. That " ^ " is the tip-off to exponentiation.

8 0

3 years ago

Read 2 more answers

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