Browsers ignore any values specified for the left or top properties under ____

Which of the following Teacher Tips would NOT be helpful when trying to select content from the Chrome Web Store? "Can be used a

gayaneshka [121]

Answer:

I'd say all of them have somewhat of a profound amount of viability and usefulness when it comes to teachers trying to find appropriate content, but "Can be used across subject areas" is not distinct enough, and is to broad/unclear, since quite obviously if teachers are looking for content for there students they will be looking for apps that are used for teaching certain subject areas. Every app used for teaching will be used for certain subject areas, so stating that statement is a mere waste of time & space. Where as "This app is not available in the Chrome Web Store" is pretty helpful to know, because knowing if an app is available or not is really important. "Helped my struggling students really understand the concept of color harmony." may be helpful if you are trying to find an app that correlates to science as the concept of color harmony is science. And "Doesn’t have much of a learning curve" shows that the app can be used long-term and is informative.

4 0

2 years ago

A circuit contains three resistors connected in parallel. The value of R1 is 2 K , the value or R2 is 6 K , and the value of R3

mezya [45]

In a parallel connection, the voltage is same in every branch.
Now, three three resistors connected in parallel.
R1 = 2k ohm
R2 = 6k ohm
R3 = 10k ohm
in parallel, net resisitance = $\frac{ R_{1}R_{2}R_{3} }{R_{1}R_{2} + R_{2}R_{3} + R_{3}R_{1}}$

Now, putting the values, we get, R net = 1.30 k ohm.
Ans, voltage = 100 VDC
Thus, power = $\frac{ V^{2} }{R} = \frac{ 100^{2} }{1.3 k} \\ \\ where \\ k = 10^{3}$
= 7.69 Watt

6 0

2 years ago

Budget Analysis (use while loop) Write a program that asks the user to enter the amount that he or she has budgeted for a month.

Agata [3.3K]

Answer:

print('This program will help you determine whether you\'ve budgeted enough ' +

'for your expenses. You\'ll just need to enter your budget and the ' +

'cost of each of your expenses for the month and we will calculate ' +

'the balance.')

budget = float(input('How much have you budgeted this month? '))

expenses = 0

check = 0

while check >= 0:

check = float(input('Enter an expense amount or -1 to quit: '))

if check != -1:

expenses += check

balance = budget - expenses

if balance < 0:

print('\nYou haven\'t budgeted enough. You\'re going to be $', \

format(-1 * balance, ',.2f'), ' short this month.', sep = '')

elif balance == 0:

print('\nBe careful. You\'ve budgeted just enough to make it through ' +

'the month.')

else:

print('\nYou will have $', format(balance, ',.2f'), ' extra this month.', \

sep = '')

6 0

1 year ago

Given the plaintext {0F0E0D0C0B0A09080706050403020100} and the key {02020202020202020202020202020202}: a. Show the original cont

Fiesta28 [93]

Answer:

Check the explanation

Explanation:

We will be arranging the plaintext in a 4 * 4 matrix (which is a rectangular shaped group of symbols, numbers, or expressions, that are arranged in columns and set rows For example, the dimension of the matrix we will be using to explain the above question is 4 × 4, because there are four rows and four columns.) in the step by step explanation below.

5 0

2 years ago

Translate the following MIPS code to C. Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3

Romashka [77]

Answer:

f = 2 * (&A[0])

See explaination for the details.

Explanation:

The registers $s0, $s1, $s2, $s3, and $s4 have values of the variables f, g, h, i, and j respectively. The register $s6 stores the base address of the array A and the register $s7 stores the base address of the array B. The given MIPS code can be converted into the C code as follows:

The first instruction addi $t0, $s6, 4 adding 4 to the base address of the array A and stores it into the register $t0.

Explanation:

If 4 is added to the base address of the array A, then it becomes the address of the second element of the array A i.e., &A[1] and address of A[1] is stored into the register $t0.

C statement:

$t0 = $s6 + 4

$t0 = &A[1]

The second instruction add $t1, $s6, $0 adding the value of the register $0 i.e., 32 0’s to the base address of the array A and stores the result into the register $t1.

Explanation:

Adding 32 0’s into the base address of the array A does not change the base address. The base address of the array i.e., &A[0] is stored into the register $t1.

C statement:

$t1 = $s6 + $0

$t1 = $s6

$t1 = &A[0]

The third instruction sw $t1, 0($t0) stores the value of the register $t1 into the memory address (0 + $t0).

Explanation:

The register $t0 has the address of the second element of the array A (A[1]) and adding 0 to this address will make it to point to the second element of the array i.e., A[1].

C statement:

($t0 + 0) = A[1]

A[1] = $t1

A[1] = &A[0]

The fourth instruction lw $t0, 0($t0) load the value at the address ($t0 + 0) into the register $t0.

Explanation:

The memory address ($t0 + 0) has the value stored at the address of the second element of the array i.e., A[1] and it is loaded into the register $t0.

C statement:

$t0 = ($t0 + 0)

$t0 = A[1]

$t0 = &A[0]

The fifth instruction add $s0, $t1, $t0 adds the value of the registers $t1 and $t0 and stores the result into the register $s0.

Explanation:

The register $s0 has the value of the variable f. The addition of the values stored in the regsters $t0 and $t1 will be assigned to the variable f.

C statement:

$s0 = $t1 + $t0

$s0 = &A[0] + &A[0]

f = 2 * (&A[0])

The final C code corresponding to the MIPS code will be f = 2 * (&A[0]) or f = 2 * A where A is the base address of the array.

3 0

1 year ago

Browsers ignore any values specified for the left or top properties under _____ positioning.