Explain the steps needed to determine the value of the expression below. 3(y-4)-7y+3/4

Helpppppppppppppppppppppppp

wariber [46]

Answer:

3.8 million lbs in tons is equivalent to 1900 pounds

Step-by-step explanation:

2000 lbs is equal to 1 ton so 3800000 divided by 2000 equals 1900.

8 0

2 years ago

Read 2 more answers

Have to bake a lot of brownies to sell! You have plenty of flour, cocoa, milk, and oil, but you only have 6 1/2 sticks of butter

Tems11 [23]

Answer: 8 + 2/3 batches of brownies you can bake.

Step-by-step explanation:

6 1/2 butter = 6 + 1/2 = (2·6 + 1)/2 = 13/2 sticks of butter, we have this

Since we have 13/2 sticks of butter and each batch of brownies needs 3/4 of a stick of butter, we have to divide the amount of butter we have between the amount of butter we use to bake a batch of brownies.

13/2 butter ÷ 3/4 butter/batch = 13·4/2·3 batches = 52/6 batches = 8 + 2/3 batches

Answer: 8 + 2/3 batches of brownies you can bake.

$\textit{\textbf{Spymore}}$

4 0

3 years ago

Twice the sum of a number and 4 is less than 12

Allisa [31]

To find the number you can set up a small equation (x being the number) and solve for x.
It can be written as 2 ( x + 4) < 12
Solve for x to find the number.
2 ( x + 4) < 12
Distribute the 2
2x+8<12
Subtract 8 on both sides
2x<4
Divide by 2 on both sides
x<2

Therefore the number is any number less than 2

~I hope this helps!~

3 0

3 years ago

The sum of the ages of Arturo, Benny, and Carlos is 41 years. Twice Arturo's age exceeds the sum of Benny and Carlos's ages by o

shepuryov [24]

Answer: Arturo is 14 years

Benny is 17 years

Carlos is 10 years

Step-by-step explanation:

Let x represent the age of Arturo.

Let y represent the age of Benny.

Let z represent the age of Carlos.

The sum of the ages of Arturo, Benny, and Carlos is 41 years. It means that

x + y + z = 41- - - - - - - - - - - - -1

Twice Arturo's age exceeds the sum of Benny and Carlos's ages by one year. It means that

2x = y + z + 1- - - - - - - - - 2

Five years ago, Benny was 2 years more than twice as old as Carlos. It means that

y - 5 = 2(z - 5) + 2

y - 5 = 2z - 10 + 2

y = 2z - 8 + 5

y = 2z - 3

Substituting y = 2z - 3 into equation 2, it becomes

2x = 2z - 3 + z + 1

2x = 2z + z - 3 + 1

2x = 3z - 2

x = 3z/2 - 1

Substituting y = 2z - 3 and

x = 3z/2 - 1 into equation 1, it becomes

3z/2 - 1 + 2z - 3 + z = 41

Multiplying through by 2, it becomes

3z - 2 + 4z - 6 + 2z = 82

3z + 4z + 2z - 2 - 6 = 82

9z - 8 = 82

9z = 82 + 8 = 90

z = 90/9

z = 10

x = 3z/2 - 1 = (3 × 10)/2 - 1

x = 14

y = 2z - 3 = 2 × 10 - 3

y = 17

6 0

4 years ago

Can someone explain step by step how to do this problem? Thanks! Calculus 2

Ganezh [65]

Answer:

1.314 MJ

Step-by-step explanation:

As water is removed from the tank, decreasing amounts are raised increasing distances. The total work done is the integral of the work done to raise an incremental volume to the required height.

There are a couple of ways this can be figured. The "easy way" involves prior knowledge of the location of the center of mass of a cone. Effectively, the work required is that necessary to raise the mass from the height of its center to the height of the discharge pipe.

The "hard way" is to write an expression for the work done to raise an incremental volume, then integrate that over the entire volume. Perhaps this is the method expected in a Calculus class.

<h3>Mass of water</h3>

The mass of the water being raised is the product of the volume of the cone and the density of water.

The cone volume is ...

V = 1/3πr²h . . . . . . for radius 2 m and height 8 m

V = 1/3π(2 m)²(8 m) = 32π/3 m³

The mass of water in the cone is then ...

M = density × volume

M = (1000 kg/m³)(32π/3 m³) ≈ 3.3510×10^4 kg

<h3>Center of mass</h3>

The center of mass of a cone is 1/4 of the distance from the base to the point. In this cone, it is (1/4)(8 m) = 2 m from the base.

The discharge pipe is 2 m above the base of the cone, so is 4 m above the center of mass. The work required to lift the mass from its center to a height of 4 m above its center is ...

W = Fd = (9.8 m/s²)(3.3510×10^4 kg)(4 m) = 1.3136×10^6 J

As the water level in the conical tank decreases, the remaining volume occupies a space that is similar to the entire cone. The scale factor is the ratio of water depth to the height of the tank: (y/8). The remaining volume is the total volume multiplied by the cube of the scale factor.

V(y) = (32π/3)(y/8)³

The differential volume at height y is the derivative of this:

dV = π/16y²

The work done to raise this volume of water to a height of 10 m is ...

(9.8 m/s²)(1000 kg/m³)(dV)((10 -y) m) = 612.5π(y²)(10 -y) J

The total work done is the integral over all heights:

$\displaystyle W=612.5\pi\int_0^8{(10y^2-y^3)}\,dy=\left.612.5\pi y^3\left(\dfrac{10}{3}-\dfrac{y}{4}\right)\right|_0^8\\\\W=612.5\pi\dfrac{2048}{3}\approx\boxed{1.3136\times10^6\quad\text{joules}}$

It takes about 1.31 MJ of work to empty the tank.

8 0

2 years ago