Hello it's my new id<br>I am numu

Each cout statement has a syntax error. Type the first cout statement, and press Run to observe the error message. Fix the error

saul85 [17]

Answer:

1. cout << "Num: " << songNum << endl;

2. cout << songNum << endl;

3. cout << songNum <<" songs" << endl;

Explanation:

//Full Code

#include <iostream>

using namespace std;

int main ()

{

int songNum;

songNum = 5;

cout << "Num: " << songNum << endl;

cout << songNum << endl;

cout << songNum <<" songs" << endl;

return 0;

}

1. The error in the first cout statement is that variable songnum is not declared.

C++ is a case sensitive programme language; it treats upper case and lower case characters differently.

Variable songNum was declared; not songnum.

2. Cout us used to print a Variable that has already been declared.

The error arises in int songNum in the second cout statement.

3. When printing more than one variables or values, they must be separated with <<

4 0

3 years ago

Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w

Dafna11 [192]

Answer:

$\mathbf{h_2 =1021.9 \ mm}$

Explanation:

Given that :

The initial volume of water $V_1$ = 1000.00 mL = 1000000 mm³

The initial temperature of the water $T_1$ = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water $T_2$ = 70° C

The coefficient of thermal expansion for the glass is ∝ = $3.8*10^{-6 } mm/mm \ per ^oC$

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature $T_1 = 10 ^ 0C$ the density of the water is $\rho = 999.7 \ kg/m^3$

At temperature $T_2 = 70^0 C$ the density of the water is $\rho = 977.8 \ kg/m^3$

The mass of the water is $\rho V = \rho _1 V_1 = \rho _2 V_2$

Thus; we can say $\rho _1 V_1 = \rho _2 V_2$ ;

⇒ $999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2$

$V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }$

$V_2 = 1022.40 \ mL$

$v_2 = 1022400 \ mm^3$

Thus, the volume of the water after heating to a required temperature of $70^0C$ is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

$V_1 = A_1 *h_1$

The area of the water before heating is:

$A_1 = \dfrac{V_1}{h_1}$

$A_1 = \dfrac{1000000}{1000}$

$A_1 = 1000 \ mm^2$

The area of the heated water is :

$A_2 = A_1 (1 + \Delta t \alpha )^2$

$A_2 = A_1 (1 + (T_2-T_1) \alpha )^2$

$A_2 = 1000 (1 + (70-10) 3.8*10^{-6} )^2$

$A_2 = 1000.5 \ mm^2$

Finally, the depth of the heated hot water is:

$h_2 = \dfrac{V_2}{A_2}$

$h_2 = \dfrac{1022400}{1000.5}$

$\mathbf{h_2 =1021.9 \ mm}$

Hence the depth of the heated hot water is $\mathbf{h_2 =1021.9 \ mm}$

4 0

3 years ago

Ann’s Retail, a women’s clothing store, hires female attendants to assist clients in the store’s dressing rooms. Larry, a male,

mojhsa [17]

Answer:

A bona fide occupational qualification defense

Explanation:

Since the store is for women clothing, the retail may prefer to employ only female to assist the customers. Under a bona fide occupational qualification defense, an employer is allowed to discriminate if a characteristic is a necessity for the performance of the job and for the business. Therefore, the store has a bona fide occupational qualification defense.

3 0

3 years ago

Is it better to do blue prints with paper and pencil or a computer program if you’re going to design a house? Why?

Hatshy [7]

Answer:

Computer program

Explanation:

I use Revit and its way better to do all that you can see 2D 3D the measurements and its super easy to use hope this helps

6 0

3 years ago

How do we find percentage error in measuring voltage across a resistor

Black_prince [1.1K]

Answer:

use the percentage error relation

Explanation:

The percentage error in anything is computed from ...

%error = ((measured value)/(accurate value) -1) × 100%

__

The difficulty with voltage measurements is that the "accurate value" may be hard to determine. It can be computed from the nominal values of circuit components, but there is no guarantee that the components actually have those values.

Likewise, the measuring device may have errors. It may or may not be calibrated against some standard, but even measurement standards have some range of possible error.

6 0

3 years ago

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Hello it's my new idI am numu ​

Hello it's my new idI am numu