Elijah buys a one-pint bottle of juice for $2.88. What is the unit rate of the cost of the juice per fluid ounce?

Calculate the potential energy associated with 1 m^3 of water at 607 feet tall taking the mass of 1 m^3 of water to be 1000 kg

Lady bird [3.3K]

Answer:

1813.137 KJ

Step-by-step explanation:

potential energy of the body = mgH

where m is mass in Kg , g= 9.81 m/sec^2 and H= height in m

here m= 1000 kg, g= 9.81 m/s^2 and H= 607 feet = 607×0.305= 185.135 m

hence the potential energy p= 1000×9.81×185.135= 1813137.2 J

= 1813.137 KJ

hence the potential energy associated with 1 m^3 of water at 607 feet tall taking the mass as 1000 kg is = 1813.137 KJ

4 0

3 years ago

Suppose there are n people on a bus. At the next stop, 9 more people get on the bus. No one gets of the bus. Draw a bar diagram

nordsb [41]

Answer:

Algebaric Expression & Bar Diagram graphic representation of number of people in bus.

Step-by-step explanation:

No. of people on bus = n

After 9 more boarding & no one deboarding the bus , no. of people = n + 9

Bar diagrams is graphical method for presentation of data. Presenting the given in bar diagram :

7 0

3 years ago

How can I find the volume of a cone 9 radius and 18 height

Anastasy [175]

Answer:

V = 486 $\pi$ , or approx. 1,526.814

Step-by-step explanation:

V = $\pi$ *r²*(h/3)

plug in the numbers: V = $\pi$ * (9)² * (18/3)

simplify: V = $\pi$ * 81 * 6

simplify: V = 486 $\pi$ , or approx. 1,526.814

6 0

3 years ago

Read 2 more answers

6x+4(150/x)=120 plz awnser for me

Gennadij [26K]

Answer:

x=10

Explaination:

4 0

3 years ago

Ayuda necesito resolver este problema con procedimiento ;)

Paraphin [41]

$x^3-2x^2+x-1$ is one of the prime factors of the polynomial

<h3>How to factor the expression?</h3>

The question implies that we determine one of the prime factors of the polynomial.

The polynomial is given as:

$x^8 - 3x^6 + x^4 - 2x^3 - 1$

Expand the polynomial by adding 0's in the form of +a - a

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^7 - 4x^6 +x^6 + 2x^5 -2x^5- 3x^4 + 4x^4 + 2x^3 -6x^3+2x^3- x^2 -3x^2 +4x^2-2x+2x-1$

Rearrange the terms

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^5 - 3x^4 + 2x^3 - x^2 + 2x^7 - 4x^6 + 4x^4 -6x^3+4x^2-2x+x^6-2x^5+2x^3-3x^2+2x-1$

Factorize the expression

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^2(x^6-2x^5+2x^3-3x^2+2x-1) + 2x(x^6-2x^5+2x^3-3x^2+2x-1) + 1(x^6-2x^5+2x^3-3x^2+2x-1)$

Factor out x^6-2x^5+2x^3-3x^2+2x-1

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x^2+2x + 1)(x^6-2x^5+2x^3-3x^2+2x-1)$

Express x^2 + 2x + 1 as a perfect square

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6-2x^5+2x^3-3x^2+2x-1)$

Expand the polynomial by adding 0's in the form of +a - a

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^4-x^3 +x^3-2x^3-x^2 -2x^2 +x+x - 1)$

Rearrange the terms

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^3-x^4-2x^3-x^2+x+x^3-2x^2 +x - 1)$

Factorize the expression

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3(x^3-2x^2+x-1) -x(x^3-2x^2+x-1)+1(x^3-2x^2+x-1))$

Factor out x^3-2x^2+x-1

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3 -x+1)(x^3-2x^2+x-1)$

One of the factors of the above polynomial is $x^3-2x^2+x-1$ .

This is the same as the option (c)

Hence, $x^3-2x^2+x-1$ is one of the prime factors of the polynomial