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3 years ago

State the number of terms for each of the following algebraic expression 2x+1

Engineering

1 answer:

harina [27]3 years ago

6 0

Answer:

Expressions are made up of terms.

A term is a product of factors.

Coefficient is the numerical factor in the term

Before moving to terms like monomials, binomials, and polynomials, like and unlike terms are discussed.

When terms have the same algebraic factors, they are like terms.

When terms have different algebraic factors, they are unlike terms.

Explanation:

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Comparison of density values determines whether an item will float or sink in water. For each of the values below, determine the

geniusboy [140]

Answer:

a) the object floats

b) the object floats

c) the object sinks

Explanation:

when an object is less dense than in the fluid in which it is immersed, it will float due to its weight and volume characteristics, so to solve this problem we must find the mass and volume of each object in order to calculate the density and compare it with that of water

volumen for a cube

V=L^3

L=1.53in=0.0388m

V=0.0388 ^3=5.8691x10^-5m^3=58.69ml

density=m/v

density=13.5g/58.69ml=0.23 g/ml

The wooden block floats because it is less dense than water

m=111mg=0.111g

density=m/v

density=0.111g/0.296ml=0.375g/ml

the metal paperclip floats because it is less dense than water

V=0.93cups=220.0271ml

m=0.88lb=399.1613g

Density=m/v

density=399.1613/220.027ml=1.8141g/ml

the apple sinks because it is denser than water

4 0

3 years ago

The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet

alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

$T_{h_i}=200^{\circ}C$

$T_{h_o}=120^{\circ}C$

$T_{c_i}=100^{\circ}C$

$T_{c_o}=125^{\circ}C$

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

$m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]$

$\frac{m_hc_{ph}}{m_cc_{pc}}$ = $\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )$

we can see that heat capacity of hot fluid is minimum

Also from energy balance

$Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )$

$NTU=\frac{UA}{\left ( mc_p\right )_{h}}$ = $\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}$

$T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}$

$T_m=41.63^{\circ}C$

NTU=1.921

$And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}$

$\epsilon =\frac{1-0.2669}{1-0.0834}$

$\epsilon =0.799$

5 0

3 years ago

A/an_ Oscilloscope uses a cathode ray tube and displays all voltages.

MrRa [10]

Answer:

I think it’s B, digital

Explanation:

3 0

3 years ago

Assume the work done compressing the He gas is -63 kJ and the internal energy change of the gas is 79 kJ. What is the heat loss

klemol [59]

Answer:

Heat gain of 142 kJ

Explanation:

We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.

Δ $U = Q + W$ ⇒ $Q =$ Δ $U -W$