All categories

3 years ago

Answer these two questions pleaseeee

Mathematics

1 answer:

Leya [2.2K]3 years ago

3 0

Answer:

1. 2/5

2.no

Step-by-step explanation:

thank you come again

You might be interested in

Why is 6/8 greater than 5/8 but less than 7/8

My name is Ann [436]

Simple!

Since we can observe that they all have the same denominators (bottom part) we can just observe the numerator (top part)

6 is indeed greater than 5, yes? So we can safely say that

6/8 > 5/8

Now the for the second part.

6 is smaller than 7, yes? So we can make the observation that

6/8 < 7/8

Hope that makes sense! (Be sure to thank me, plz)

P.S. If you want an example of unlike denominators, let me know. <span />

7 0

3 years ago

The balance in your bank account is $720.50. You use your debit card for purchases of $45, $82, and $22.25. Then, you deposit $9

myrzilka [38]

Answer:

669 dollars and 70 cents

Step-by-step explanation:

just minus all the money spent and add the money added

7 0

3 years ago

Can anyone help me with this please ?

zlopas [31]

$\displaystyle \bf\\
 \frac{x}{20} = \frac{14}{25} \\ \\ \\ 
x = \frac{14 \cdot 20}{25} = \frac{280}{25} = 11,2 \sim \boxed{10} ~~\text{(rounded to the nearest tenth)}$

7 0

3 years ago

Find the value of y.<br> (6x - 13)<br> (3y + 1)<br> (8x - 61)

Korolek [52]

Answer:

y=1/3

Step-by-step explanation:

3y+1=0

3y=-1

3y/3=-1/3

y=-1/3

6 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