How do Consumers sometimes interact with a producers?

Assuming that the following three variables have already been declared, which variable will store a Boolean value after these st

Yuki888 [10]

Answer:

C

Explanation:

Boolean Algebra deals with either a one or a zero and how to manipulate them in computers or elsewhere. The "choice" option may not work, since for text it must be enclosed in quotation marks, usually. For "again," it's text and not a 1 or 0. So, the answer is C, since this is a 0.

5 0

2 years ago

one number is 11 more than another number. find the two number if three times the larger number exceeds four times the smaller n

vaieri [72.5K]

Answer:

a = 40

b = 29

Explanation:

Give a place holder for the numbers that we don't know.

Lets call the two numbers a and b.

From the given info, we can write an expression and solve it:

"one number is 11 more than another number"

a = 11 + b

from this, we know that a > b.

''three times the larger number exceeds four times the smaller number by 4"

3a = 4b + 4

Now we have 2 equations, we can use them to solve using whatever method you want.

a = 11 + b

3a = 4b + 4

I will be using matrices RREF to solve for this.

a - b = 11

3a - 4b = 4

$\begin{bmatrix}1 & -1 & 11\\3 & -4 & 4 \end{bmatrix}$

$\begin{bmatrix}1 & 0 & 40\\0 & 1 & 29 \end{bmatrix}$

a = 40

b = 29

6 0

2 years ago

While there are many ways to solve this problem, one strategy is to calculate the volume of any metal's unit cell given its theo

IgorC [24]

Answer:

$V_{c}=1.396x10^{-28} m^{3} /unit.cell$

Explanation:

z = number of atoms

M = Molar mass of zirconium

N = Avogadro’s number

Vc = volume of zirconium unit cell

d = density

$z=12x\frac{1}{6}+2x\frac{1}{2}+3=6$

z = 6 atoms per unit cell

M = 91.224 g/mol

N = $6.023x10^{23} atoms/mol$

d = $6.51g/cm^{3}$

$V_{c}=\frac{zxM}{dxN}$

$V_{c}=\frac{6x(91.224g/mol)}{(6.51g/cm^{3}) x(6.023x10^{23}atoms/mol) }$

$V_{c}=1.396x10^{-22} cm^{3} /unit.cell$

$V_{c}=1.396x10^{-28} m^{3} /unit.cell$

8 0

2 years ago

A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s)

4vir4ik [10]

A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s) with a flow rate of 8.7 m^3/s, 6 mg/L BOD5 and 8.3 mg/L DO. Both streams are at 20Â°C. After mixing, the river is 3 meters deep and flowing at a velocity of 0.50 m/s. DOsat for this river is 9.0 mg/L. The deoxygenation constant is kd= 0.20 d^-1 and The reaction rate constant k at 20 Â°C is 0.27 d^-1.

The answer therefore would be the number 0.27 divided by two and then square while getting the square you would make it a binomial.

I wont give the answer but the steps

Your Welcome

8 0

3 years ago

Liquid benzene and liquid n-hexane are blended to form a stream flowing at a rate of 1700 lbm/h. An on-line densitometer (an ins

Taya2010 [7]

Let me think of that

5 0

2 years ago