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3 years ago

7 features of QBASIC

1 answer:

7 0

A. QBASIC does not use technical terminology (word) to write statements.
b. QBASIC automatically checks syntax.
c. QBASIC capitalizes the reserved words.
d. QBASIC keeps the same variable name used in a program to identical form.
e. QBASIC allows you to break lengthy programs into modules.
<span>f. QBASIC interprets a statement of a program at a time to CPU</span>

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Read the scenario and answer the question:

mel-nik [20]

C even the rest of answers seem practical c would be most likely

4 0

2 years ago

Read 2 more answers

Note: You can use a word document to write your answers and copy-paste your answer to the area specified. a. (5 points) Convert

MrMuchimi

Answer:

$EA9_{16} = 3753$

$CB2_{16} = 3250$

$(1011 1110 1101 1011 1010)_2 = 781754$

$(1010 1000 1011 1000 1110 1101)_2 = 11057389$

$(1011 1110 1101 1011 1010)_2 = BEDBA$

$(1010 1000 1011 1000 1110 1101)_2 = A8B8ED$

$74510_8= 221416$

$67210_8 = 203212$

Explanation:

Solving (a): To base 10

$(i)\ EA9_{16$

We simply multiply each digit by a base of 16 to the power of their position.

i.e.

$EA9_{16} = E * 16^2 + A * 16^1 + 9 * 16^0$

$EA9_{16} = E * 256 + A * 16 + 9 * 1$

In hexadecimal

$A = 10; E = 14$

So:

$EA9_{16} = 14 * 256 + 10 * 16 + 9 * 1$

$EA9_{16} = 3753$

$(ii)\ CB2_{16}$

This gives:

$CB2_{16} = C * 16^2 + B * 16^1 + 2 * 16^0$

$CB2_{16} = C * 256 + B * 16 + 2 * 1$

In hexadecimal

$C = 12; B =11$

So:

$CB2_{16} = 12 * 256 + 11 * 16 + 2 * 1$

$CB2_{16} = 3250$

Solving (b): To base 10

$(i)\ (1011 1110 1101 1011 1010)_2$

We simply multiply each digit by a base of 2 to the power of their position.

i.e.

$(1011 1110 1101 1011 1010)_2 = 1 * 2^{19} + 0 * 2^{18} + 1 * 2^{17} + 1 * 2^{16} +1 * 2^{15} + 1 * 2^{14} + 1 * 2^{13} + 0 * 2^{12} + 1 * 2^{11} + 1 * 2^{10} + 0 * 2^9 + 1 * 2^8 +1 * 2^7 + 0 * 2^6 + 1 * 2^5 + 1 * 2^4 + 1 * 2^3 + 0 * 2^2 + 1 * 2^1 + 0 * 2^0$

$(1011 1110 1101 1011 1010)_2 = 781754$

$(ii)\ (1010 1000 1011 1000 1110 1101)_2$

$(1010 1000 1011 1000 1110 1101)_2 = 1 * 2^{23} + 0 * 2^{22} + 1 * 2^{21} + 0 * 2^{20} +1 * 2^{19} + 0 * 2^{18} + 0 * 2^{17} + 0 * 2^{16} + 1 * 2^{15} + 0 * 2^{14} + 1 * 2^{13} + 1 * 2^{12} +1 * 2^{11} + 0 * 2^{10} + 0 * 2^9 + 0 * 2^8 + 1 * 2^7 + 1 * 2^6 + 1 * 2^5 + 0 * 2^4 + 1*2^3 + 1 * 2^2 + 0 * 2^1 + 1 * 2^0$

$(1010 1000 1011 1000 1110 1101)_2 = 11057389$

Solving (c): To base 16

$i.\ (1011 1110 1101 1011 1010)_2$

First, convert to base 10

In (b)

$(1011 1110 1101 1011 1010)_2 = 781754$

Next, is to divide 781754 by 16 and keep track of the remainder

$781754/16\ |\ 48859\ R\ 10$

$48859/16\ |\ 3053\ R\ 11$

$3053/16\ |\ 190\ R\ 13$

$190/16\ |\ 11\ R\ 14$

$11/16\ |\ 0\ R\ 11$

Write out the remainder from bottom to top

$(11)(14)(13)(11)(10)$

In hexadecimal

$A = 10; B = 11; C = 12; D = 13; E = 14; F = 15.$

$(11)(14)(13)(11)(10)=BEDBA$

So:

$(1011 1110 1101 1011 1010)_2 = BEDBA$

$ii.\ (1010 1000 1011 1000 1110 1101)_2$

In b

$(1010 1000 1011 1000 1110 1101)_2 = 11057389$

Next, is to divide 11057389 by 16 and keep track of the remainder

$11057389/16\ |\ 691086\ R\ 13$

$691086/16\ |\ 43192\ R\ 14$

$43192/16\ |\ 2699\ R\ 8$

$2699/16\ |\ 168\ R\ 11$

$168/16\ |\ 10\ R\ 8$

$10/16\ |\ 0\ R\ 10$

Write out the remainder from bottom to top

$(10)8(11)8(14)(13)$

In hexadecimal

$A = 10; B = 11; C = 12; D = 13; E = 14; F = 15.$

$(10)8(11)8(14)(13) = A8B8ED$

So:

$(1010 1000 1011 1000 1110 1101)_2 = A8B8ED$

Solving (d): To octal

$(i.)\ 74510$

Divide 74510 by 8 and keep track of the remainder

$74510/8\ |\ 9313\ R\ 6$

$9313/8\ |\ 1164\ R\ 1$

$1164/8\ |\ 145\ R\ 4$

$145/8\ |\ 18\ R\ 1$

$18/8\ |\ 2\ R\ 2$

$2/8\ |\ 0\ R\ 2$

Write out the remainder from bottom to top

$74510_8= 221416$

$(ii.)\ 67210$

Divide 67210 by 8 and keep track of the remainder

$67210/8\ |\ 8401\ R\ 2$

$8401/8\ |\ 1050\ R\ 1$

$1050/8\ |\ 131\ R\ 2$

$131/8\ |\ 16\ R\ 3$

$16/8\ |\ 2\ R\ 0$

$2/8\ |\ 0\ R\ 2$

Write out the remainder from bottom to top

$67210_8 = 203212$

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