Write 5 chart types

What does choosing Slide Sorter view do?

elixir [45]

Answer:

The last two line i.e "displays miniature versions of all the presentation's slides " is the correct answer to the given question .

Explanation:

The main objective of the slide sorter is showing the miniature versions of all the slides though user can quickly push the slides and organize the slides in well mannered by moving the slides . The Slide Sorter display gave the viewers to thumbnail the overview of the slides.

The slide sorter making the presentation effective by organizing and sorting the slide .
All the other option are not correct for the Slide Sorter view that's why these are incorrect option .

3 0

3 years ago

For a new version of processor, suppose the capacitive load remains, how much more energy will the processor consume if we incre

erastovalidia [21]

Answer:

The answer is below

Explanation:

The amount of power dissipated by a processor is given by the formula:

P = fCV²

Where f = clock rate, C = capacitance and V = Voltage

For the old version of processor with a clock rate of f, capacitance C and voltage of V, the power dissipated is:

P(old) = fCV²

For the new version of processor with a clock rate of 20% increase = (100% + 20%)f = 1.2f, capacitance is the same = C and voltage of 20% increase = 1.2V, the power dissipated is:

P(new) = 1.2f × C × (1.2V)² = 1.2f × C × 1.44V² =1.728fCV² = 1.728 × Power dissipated by old processor

Hence, the new processor is 1.728 times (72.8% more) the power of the old processor

4 0

3 years ago

Anyone use zoom<br><br>code:- 2574030731<br>pass:- HELLO<br>Z●●M

r-ruslan [8.4K]

Answer:

ok be there in a sec

Explanation:

3 0

3 years ago

Read 2 more answers

Would this program compile run? If not, why; if yes: what would be the output? public class ConstChaini { public static void mai

nikdorinn [45]

Answer:

Check the explanation

Explanation:

// Code to copy

public class ConstChain1 {

public static void main(String[] args)

{

new SubClass();

System.out.println();

new SubClass(1);

}

class SuperClass{

public SuperClass() {

System.out.println("D");

}

public SuperClass(int i) {

System.out.println("C");

}

class SubClass extends SuperClass{

public SubClass() {

this(10);

System.out.println("B");

}

public SubClass(int i) {

super(i);

System.out.println("A");

}

Explanation:

Now each time when a object of a subclass is been created it's super class constructor will be executed first then only subclass object is created by executing it's constructor.

8 0

3 years ago

(Assignment 1, individual) Create proc3.s Study the proc3.c and re-write the same program in MIPS with the following requirement

Vedmedyk [2.9K]

Answer:

text

.globl main

main:

li $s0,5 #load 5 to x

li $s1,10 #load 10 to y

move $a0,$s0

move $a1,$s1 #passing argument to sum function

jal sum

add $s1,$s1,$v0 #get y + sum(x,y)

add $s1,$s1,$s0 #get x+ y + sum(x,y)

li $v0,1

move $a0,$s1

syscall #print value of y

li $v0,10 #terminate call

syscall

sum:

addi $sp, $sp, 4

subu $sp,$sp,4 # point to the place for the new item,

sw $ra,($sp) # store the contents of $ra as the new top.

move $t1, $a0 #store parameters m

move $t2, $a1 #store parameters n

add $a0,$t2,1 #get n+1

add $a1,$t1,1 #get m+1

jal sub

move $t3,$v0 #store result to t3

sub $a0,$t1,1 #get m-1

sub $a1,$t2,1 #get n-1

jal sub

move $t4,$v0 #store result to t3

add $v0,$t3,$t4 #return p+q

lw $ra,($sp) # store the contents of $ra as the new top.

addu $sp,$sp,4 # point to the place for the new item,

addi $sp, $sp, 4

jr $ra

sub:

sub $v0,$a1,$a0 #return b-a

jr $ra

Explanation:

text

.globl main

main:

li $s0,5 #load 5 to x

li $s1,10 #load 10 to y

move $a0,$s0

move $a1,$s1 #passing argument to sum function

jal sum

add $s1,$s1,$v0 #get y + sum(x,y)

add $s1,$s1,$s0 #get x+ y + sum(x,y)

li $v0,1

move $a0,$s1

syscall #print value of y

li $v0,10 #terminate call

syscall

sum:

addi $sp, $sp, 4

subu $sp,$sp,4 # point to the place for the new item,

sw $ra,($sp) # store the contents of $ra as the new top.

move $t1, $a0 #store parameters m

move $t2, $a1 #store parameters n

add $a0,$t2,1 #get n+1

add $a1,$t1,1 #get m+1

jal sub

move $t3,$v0 #store result to t3

sub $a0,$t1,1 #get m-1

sub $a1,$t2,1 #get n-1

jal sub

move $t4,$v0 #store result to t3

add $v0,$t3,$t4 #return p+q

lw $ra,($sp) # store the contents of $ra as the new top.

addu $sp,$sp,4 # point to the place for the new item,

addi $sp, $sp, 4

jr $ra

sub:

sub $v0,$a1,$a0 #return b-a

jr $ra

The above program takes in Local variables mapping: main(): x -> $s0, y -> $s1 b. sum(): p -> $s0, q -> $s1 Then Input arguments mappings: sum(): m -> $a0, n-> $a1 b. sub(): a -> $a0, b-> $a1

And return all values from a function which must be stored in V registers in ascending order.

8 0

3 years ago

Write 5 chart types​

Write 5 chart types