Describe a way that someone could estimate a length of a book

The regular price of an item at a store is p dollars. The item is on sale for 20% off the regular price. Some of the expressions

s344n2d4d5 [400]

Two expressions each represent the sale price of the item are:

Expression B : 0.8p is correct

Expression D: p - 0.2p is correct

<em><u>Solution:</u></em>

The regular price of an item at a store is p dollars

The item is on sale for 20% off the regular price

Therefore,

Regular price = p dollars

Discount = 20 % of regular price

Therefore,

Discount = 20 % of p

$discount = 20 \% \times p\\\\discount = \frac{20}{100} \times p = 0.2 \times p\\\\discount = 0.2p$

Sale price = Regular price - discount price

Sale price = p - 0.2p

Thus Expression D: p - 0.2p is correct

On solving, sale price = p - 0.2p

Sale price = 0.8p

Thus expression B : 0.8p is correct

3 0

3 years ago

Find The Area Of This Shape

spin [16.1K]

14 because you count the height and how much the base is from the farthest point left to right and multiply h=2 b=7 therefor 2x7 is 14

7 0

3 years ago

Maggie has a food scale with a digital display that shows the weight as a decimal number. The recipe she is using calls for ingr

strojnjashka [21]

Put the decimal over 100 and get rid of the decimal

8 0

3 years ago

A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and

nasty-shy [4]

Answer:

correct option is C) 2.8

Step-by-step explanation:

given data

string vibrates form = 8 loops

in water loop formed = 10 loops

solution

we consider mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = $\rho$

case (1)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

so here

$l = \frac{8 \lambda _1}{2}$ ..............1

$l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}$

and we know velocity is express as

velocity = frequency × wavelength .....................2

$\sqrt{\frac{Tension}{mass\ per\ unit \length }}$ = f × $\lambda_1$

here tension = mg

so

$\sqrt{\frac{mg}{\mu}}$ = f × $\lambda_1$ ..........................3

and

case (2)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

$l = \frac{10 \lambda _1}{2}$ ..............4

$l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}$

when block is immersed

equilibrium eq will be

Tenion + force of buoyancy = mg

T + v × $\rho$ × g = mg

and

T = v × $\rho$ - v × $\rho$ × g

from equation 2

f × $\lambda_2$ = f × $\frac{1}{5}$

$\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}$ .......................5

now we divide eq 5 by the eq 3

$\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}$

solve irt we get

$1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}$

so

relative density $\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}$

relative density = 2.78 ≈ 2.8

so correct option is C) 2.8

3 0

3 years ago

Some of the students three scores is 231. If the first is 20 points more than the second, and the sum of the first two is 6 more

Aleks [24]

Answer:

The first score is 109

Step-by-step explanation:

I am assuming that in the first sentence of the question, you meant:

Sum of the students three scores is 231...

First, let the scores of the first second and third student be a, b and c respectively. We are told that:

a + b + c = 231 . . . . . . . .(1) (sum of students three scores is 231)

a = b + 20 . . . . . . . . . . . (2) (the first is 20 points more than the second)

a + b = 6c . . . . . . . . . . . .(3) (sum of the first two is 6 more times the third)

required, find a.

substituting the value of (a + b) in equation (3) into equation (1), we will have the following:

since a + b = 6c . . . (3)

a + b + c = 231 . . . . . (1), becomes,

(a + b) + c = 231

(6c) + c = 231

7c = 231 (divide both sides by 7)

c = 231 ÷ 7 = 33

∴ c = 33

Next, from equation (2), we know that a = b + 20; this can also be written as:

a - 20 = b

∴ b = a - 20 . . . . . . . (4)

Finally, putting the value of b in equation (4) and the value of c calculated above into equation 1, ( a + b + c = 231), we have the following:

a + (a - 20) + 33 = 231

(a + a) - 20 + 33 = 231

2a + 13 = 231

2a = 231 - 13 = 218

a = 218 ÷ 2 = 109

∴ a = 109

we can also calculate for 'b' by substituting for the value of 'a' in equation 4

b = a - 20 = 109 - 20 = 89.

and to test if the values of a, b and c are correct:

a + b + c = 231

109 + 89 + 33 = 231

4 0

3 years ago