Find the subgroup of d4 and the normal and non normal subgroups of d3 and d4 using u and v, u being the flips and v being the rotations.

A non current asset is classified as held for sale. On the date of classification, immediately prior to the transfer to the 'held for sale' classification, the asset had: a cost of $100 000, accumulated depreciation of $40 000(10% per annum, straight line over 4 years); and accumulated impairment losses in terms of IAS 36 of $15 000. The asset was then impaired in terms of IFRS 5 by $10 000. Asume that the above asset had not yet been sold at the end of the following reporting date, at which point its fair value less cost to sell was $75 000.

A non current asset is classified as held for sale. On the date of classification, immediately prior to the transfer to the 'held for sale' classification, the asset had: a cost of $100 000, accumulated depreciation of $40 000(10% per annum, straight line over 4 years); and accumulated impairment losses in terms of IAS 36 of $15 000. The asset was then impaired in terms of IFRS 5 by $10 000. Asume that the above asset had not yet been sold at the end of the following reporting date, at which point its fair value less cost to sell was $75 000. a) impairment loss reversal will be $15 000. b) impairment loss reversal will be $25 000. c) impaiment loss reversal will be $40 000. d) impairment loss reversal will be $10 000.

- hierarchical or partitional
- overlapping or non-overlapping
- fuzzy or crisp
- complete or incomplete

Note: Each part should be labeled with four characteristics, e.g., partitional, overlapping, crisp, and incomplete. Also, if you feel there may be some ambiguity about what characteristics a grouping has, provide a short justification of your answer.

Case 1: The objects are the students in a class. There are groups for each official grade students received for the class.

Case 2: The objects are cities. There are groups of cities corresponding to various locations, namely, county (local region), state or province, and country.

Case 3: The objects are the applicants to a college. Each applicant is assigned a score from 0 to 10 indicating the likelihood/desirability of their admission. Even before any decisions have been made, the admissions personnel view the students as belonging to two groups: those that will be accepted and those that will be rejected.

1. A receptor tyrosine kinase (RTK) has a non-functioning kinase domain

Enhanced or Reduced

2. A G-protein whose α-subunit has a higher than normal binding affinity for GTP and a normal binding affinity for GDP.

Enhanced or Reduced

3. Protein Kinase A whose regulatory subunits cannot bind cAMP

Enhanced or Reduced

4. IP3 cannot bind to ER-calcium channels

Enhanced or Reduced

5. Notch cannot be cleaved following binding to Delta

Enhanced or Reduced

6. Sodium orthovanadate, a phosphatase inhibitor, is added to cells that have been stimulated with EGF

Enhanced or Reduced

7. Adenylyl cyclase cannot bind ATP

Enhanced or Reduced

8. GAP is hyperactivated

Enhanced or Reduced

If a non-competitive inhibitor binds to E, is it possible that it also binds to ES in many cases? Group of answer choices

a. Yes

b. No

The three steps involved in the non-constant growth dividend discount model are:
      
A. Step 1: Set the investment horizon (year H) as the future year after which you expect the company's growth to settle down to a stable rate.
      
B. Step 2: Forecast the stock price at the horizon, and discount it also to give its present value today.
      
C. Step 1: Price estimated using the constant- growth formula to value the dividends that will be paid after the horizon date.
      
D. Step 3: Sum the total present value of dividends plus the present value of the ending stock price.
      
E. Step 2: Calculate the present value of dividends from the end of investment till horizon year.
      
F. Step 3: Estimate the rate of return of the stock to compare it with the price.

Choose the right three Steps in order.

. In the space to the right of each group write NV for non-vascular or SL for seedless vascular (1 point each). List the characteristics of the sporophytes in your drawings. In particular, note a) their colors and sizes of sporophytes b) the location and colors of the sporangia; c) the presence or absence of stomata; and d) type dependency between the sporophyte and gametophyte—explained. (total 60 points)

Assignment Instructions:

1) The Factorial The factorial of a non-negative integer ??, denoted by ??!, is the product of all positive integers less than or equal to ??. The textbook has an example of a recursive MIPS implementation of factorial. Additionally, a simplified version of the MIPS assembly language recursive implementation of the factorial function is attached. Trace the factorial example carefully using QTSPIM

2) Recursive definition of multiplication The function ??????????(??, ??) for two positive integers 1 ? ??, and 1 ? ??, is defined as the following: ??????????(??, 1) = ??; ??????????(??, ??) = ?? + ??????????(??, ?? ? 1) Write a recursive version of ??????????() in C or C++ and a pseudo C program (based on chapter 2 in the book) then use these programs to develop a MIPS program that gets as input two integers 0 < ?? ? 255, and 0 < ?? ? 255, and returns the result of ??????????(??, ??) in $v1. Your deliverable should be the pseudo C and the assembly level function

Given code file:

#####################################################################################

# Functional Description: Main program to test Factorial function # Enter a negative number to terminate run

#####################################################################################

.data

.align 2

prompt:   .asciiz "\n\n Give me a value for \"N\" : "

msg: .asciiz " N factorial is: "

bye: .asciiz " \n *** Good-Bye ***"

.text

main: addiu $sp, $sp, -8 #Allocate space

mloop:

li $v0, 4

la $a0, prompt

syscall

li $v0, 5 #Get value for N

syscall

bltz $v0, quit

sw $v0, 0 ($sp)

jal Fac # Call factorial

li $v0, 4 # Print message

la $a0, msg

syscall

lw $a0, 4($sp) #Get result

li $v0, 1

syscall #Print factorial

b mloop

quit:

addiu $sp, 8 # Deallocate space

li $v0, 4 la $a0, bye

syscall li $v0, 10

syscall

#####################################################################################

# Functional Description: Recursive Factorial Fac (N: in, N! :out)

#####################################################################################

Fac:

lw $a0, 0 ($sp)

bltz $a0, Problem

addi $t1, $a0, -13

bgtz $t1, Problem # 13 is largest value we can

# accept

addiu $sp, $sp, -16 # Allocate

sw $ra, 12 ($sp) # Save return address

sw $a0, 8($sp)

slti $t0, $a0, 2 # If N is 1 or 0, then return the value 1

beqz $t0, Go

li $v0, 1

b facret

Go:

addi $a0, $a0, -1 #

sw $a0, 0 ($sp) # Pass N-1 to factorial function

jal Fac # Recursive call

lw $v0, 4($sp) # Get (N-1) ! back.

lw $ra, 12 ($sp)

lw $a0, 8 ($sp)

mult $v0, $a0 # N* (N-1) !

mflo $v0

facret:

addiu $sp, $sp, 16 # Deallocate

sw $v0, 4 ($sp)

jr $ra

Problem:

sw $0, 4 ($sp)

jr $ra

Second give code file:

#####################################################################################

# Functional Description: Main program to test Factorial function # Enter a negative number to terminate run

#####################################################################################

.data

.align 2

.text

main: addiu $sp, $sp, -8 # Allocate space

mloop:

li $v0, 4 # Get value for N

sw $v0, 0 ($sp)

jal Fac # Call factorial

or $v1, $v0, $0

addiu $sp, 8 # Deallocate space

li $v0, 10

syscall

#####################################################################################

# Functional Description: Recursive Factorial Fac (N: in, N! :out)

#####################################################################################

Fac:

lw $a0, 0 ($sp)

addiu $sp, $sp, -16 # Allocate

sw $ra, 12 ($sp) # Save return address

sw $a0, 8($sp)

slti $t0, $a0, 2 # If N is 1 or 0, then return the value 1

eqz $t0, Go

li $v0, 1

b facret

Go:

addi $a0, $a0, -1 #

sw $a0, 0 ($sp) # Pass N-1 to factorial function

jal Fac # Recursive call

lw $v0, 4($sp) # Get (N-1) ! back.

lw $ra, 12 ($sp)

lw $a0, 8 ($sp)

mult $v0, $a0 # N* (N-1) !

mflo $v0

facret:

addiu $sp, $sp, 16 # Deallocate

sw $v0, 4 ($sp)

jr $ra