Answer: I think its true but I'm not like 100% sure sorry
Explanation: i dont know just seems right
The three options that describe the typographic hierarchy are given below:
- The importance of information is shown through the text.
- The presentation of the text in terms of structure.
- The placement of the text defines its flow and hierarchy.
Thus, the correct options are A, B, and C.
<h3>What is Typographic hierarchy?</h3>
Typographic hierarchy may be characterized as an approach that utilizes typography: the size, font, and layout of distinct text components to produce a hierarchical division that can authenticate users where to look for specific information.
The purpose of using this methodology is to focus on the main points of the article accordingly. It increases the sense of understanding of the readers with focus.
Therefore, it is well described above.
To learn more about Typographic hierarchy, refer to the link:
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The two best practices for creating ads are:
- Include three to five ads and at least three extensions in each ad group.
- Optimize the campaign’s ad rotation for clicks or conversion actions.
<h3>How do one Create one responsive search ad per ad group?</h3>
In any ad group, one need to have at least a single responsive search ad.
Note that this will help to optimize one's performance, and as such, the two best practices for creating ads are:
- Include three to five ads and at least three extensions in each ad group.
- Optimize the campaign’s ad rotation for clicks or conversion actions.
Learn more about ads from
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Answer:
a) AL will contains 0011 1100
Explanation:
In assembly language, shifting bits in registers is a common and important practice. One of the shifting operations is the SHR AL, x where the x specifies that the bits be shifted to the right by x places.
SHR AL, 2 therefore means that the bits contained in the AL should be shifted to the right by two (2) places.
For example, if the AL contains binary 1000 1111, the SHR AL, 2 operation will cause the following to happen
Original bit => | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
Shift once to the right => | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | (0) |
Shift once to the right => | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | (0) | (0) |
Notice;
(i) that there are two shifts - one at a time.
(ii) that the bits in bold face are the bits in the AL after the shift. Those that in regular face are those in the carry flag.
(iii) that the new bits added to the AL after a shift are the ones in bracket. They are always set to 0.
I = V/R. So the answer is 9/200 = 0.045A = 45mA which is B
