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3 years ago

8

After adding an image to her flyer, Danica played around to see which layout would look the best. At one point, her text was on

top of her image.
Which features can Danica use to fix the issue? Check all that apply.
please help me
I ave been waiting for an hour

Position feature

Wrap text feature

Picture dialog box

Insert tab

Picture tools
can someone please help me

Computers and Technology

2 answers:

maw [93]3 years ago

7 0

Answer:

Position Feature

Wrap text feature

Picture tools

Explanation:

Norma-Jean [14]3 years ago

3 0

Answer:

Position Feature

Wrap text feature

Picture tools

Explanation:

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The soft-tissue lining of oysters and clams creates pearls. Is a pearl a mineral?

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Initialize the list short_names with strings 'Gus', 'Bob', and 'Zoe'. Sample output for the given program:

ohaa [14]

Answer:

Following are the correct code to this question:

short_names=['Gus','Bob','Zoe']#defining a list short_names that holds string value

print (short_names[0])#print list first element value

print (short_names[1])#print list second element value

print (short_names[2])#print list third element value

Output:

Gus

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Zoe

Explanation:

In the above python program code, a list "short_names" list is declared, that holds three variable that is "Gus, Bob, and Zoe".
In the next step, the print method is used that prints list element value.
In this program, we use the list, which is similar to an array, and both elements index value starting from the 0, that's why in this code we print "0,1, and 2" element value.

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3 years ago

Increasingly patients are creating and maintaining personal health records (PHRs) with data from a variety of healthcare provide

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2 years ago

(a) How many locations of memory can you address with 12-bit memory address? (b) How many bits are required to address a 2-Mega-

I am Lyosha [343]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Let,

The address of 1-bit memory to add in 2 location:

$\to \frac{0}{1} =2^1 \ (\frac{m}{m} \ location)$

The address of 2-bit memory to add in 4 location:

$\to \frac{\frac{00}{01}}{\frac{10}{11}} =2^2 \ (\frac{m}{m} \ location)$

similarly,

Complete 'n'-bit memory address' location number is = $2^n.$ Here, 12-bit memory address, i.e. n = 12, hence the numeral. of the addressable locations of the memory:

$= 2^n \\\\ = 2^{12} \\\\ = 4096$

In point b:

$\to Let \ Mega= 10^6$

$=10^3\times 10^3\\\\= 2^{10} \times 2^{10}$

So,

$\to 2 \ Mega =2 \times 2^{20}$

$= 2^1 \times 2^{20}\\\\= 2^{21}$

The memory position for ' $2^n$ ' could be 'n' m bits'

It can use $2^{21}$ bits to address the memory location of 21.

That is to say, the 2-mega-location memory needs '21' bits.

Memory Length = 21 bit Address

In point c:

$i^{th}$ element array addresses are given by:

$\to address [i] = B+w \times (i-LB)$

$_{where}, \\\\B = \text {Base address}\\w= \text{size of the element}\\L B = \text{lower array bound}$

$\to B=\$ 52\\\to w= 4 byte\\ \to L B= 0\\\to address = 10$

$\to address [10] = \$ 52 + 4 \times (10-0)\\$

$= \$ 52 + 40 \ bytes\\$

1 term is 4 bytes in 'MIPS,' that is:

$= \$ 52 + 10 \ words\\\\ = \$ 512$

In point d:

$\to base \ address = \$ t 5$

When MIPS is 1 word which equals to 32 bit :

In Unicode, its value is = 2 byte

In ASCII code its value is = 1 byte

both sizes are < 4 byte

Calculating address:

$\to address [5] = \$ t5 + 4 \times (5-0)\\$

$= \$ t5 + 4 \times 5\\ \\ = \$ t5 + 20 \\\\= \$ t5 + 20 \ bytes \\\\= \$ t5 + 5 \ words \\\\= \$ t 10 \ words \\\\$

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3 years ago