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3 years ago

Pressing the e key while in edit mode will exit the interaction mode true false

Computers and Technology

1 answer:

lara [203]3 years ago

8 0

Answer:

While in Edit mode pressing EKEY will not exit the interaction mode.

Answer : FALSE

EKEY : It is Extrude Selected. When Extrude is used in EditMode. It will transforms all the selected edges to faces.

Q key is used to Quit the Blender.

ESC Key is used to cancel any blender functions.

Explanation:

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Production cost can be reduced using technology. Explain

Black_prince [1.1K]

Explanation:

By reducing the time the use on unnecessary tasks. And you would be able to operate on using your time on beneficial tasks

3 0

3 years ago

Using the college registration example from section 6.7.3 as a starting point, do the following:

Arada [10]

Answer:

Explanation:

NCLUDE Irvine32.inc

TRUE = 1

FALSE = 0

.data

gradeAverage WORD ?

credits WORD ?

oKToRegister BYTE ?

str1 BYTE "Error: Credits must be between 1 and 30" , 0dh,0ah,0

main PROC

call CheckRegs

exit

main ENDP

CheckRegs PROC

push edx

mov OkToRegister,FALSE

; Check credits for valid range 1-30

cmp credits,1 ; credits < 1?

jb E1

cmp credits,30 ; credits > 30?

ja E1

jmp L1 ; credits are ok

; Display error message: credits out of range

E1:

mov edx,OFFSET str1

call WriteString

jmp L4

L1:

cmp gradeAverage,350 ; if gradeAverage > 350

jna L2

mov OkToRegister,TRUE ; OkToRegister = TRUE

jmp L4

L2:

cmp gradeAverage,250 ; elseif gradeAverage > 250

jna L3

cmp credits,16 ; && credits <= 16

jnbe L3

mov OkToRegister,TRUE ; OKToRegister = TRUE

jmp L4

L3:

cmp credits,12 ; elseif credits <= 12

ja L4

mov OkToRegister,TRUE ; OKToRegister = TRUE

L4:

pop edx ; endif

ret

CheckRegs ENDP

END main

8 0

3 years ago

Long Answer Questions: a. Briefly explain the types of Control Structures in QBASIC.

Alexus [3.1K]

Answer:

The three basic types of control structures are sequential, selection and iteration. They can be combined in any way to solve a specified problem. Sequential is the default control structure, statements are executed line by line in the order in which they appear.

7 0

3 years ago

Given the value x=false ,y=5 and z=1 what will be the value of F=(4%2)+2*Y +6/2 +(z&&x)?

Bezzdna [24]

<h2>Answer:</h2>

F = 13

<h2>Explanation:</h2>

Given:

x = false

y = 5

z = 1

F = (4%2)+2*Y +6/2 +(z&&x)

We solve this arithmetic using the order of precedence:

<em>i. Solve the brackets first</em>

=> (4 % 2)

This means 4 modulus 2. This is the result of the remainder when 4 is divided by 2. Since there is no remainder when 4 is divided by 2, then

4 % 2 = 0

=> (z && x)

This means (1 && false). This is the result of using the AND operator. Remember that && means AND operator. This will return false (or 0) if one or both operands are false. It will return true (or 1) if both operands are true.

In this case since the right operand is a false, the result will be 0. i.e

(z && x) = (1 && false) = 0

<em>ii. Solve either the multiplication or division next whichever one comes first.</em>

=> 2 * y

This means the product of 2 and y ( = 5). This will give;

2 * y = 2 * 5 = 10

=> 6 / 2

This means the quotient of 6 and 2. This will give;

6 / 2 = 3

<em>iii. Now solve the addition by first substituting the values calculated earlier back into F.</em>

F = (4%2)+2*Y +6/2 +(z&&x)

F = 0 + 10 + 3 + 0

F = 13

Therefore, the value of F is 13

5 0

3 years ago

Change the shape fill color to Dark Red. It is the first option in the Standard Colors section of the color palette.

maxonik [38]

Is this a question ?

3 0

3 years ago