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Naddik [55]
3 years ago
11

Select the creative imaging fields that require knowledge of programming.

Engineering
1 answer:
denis23 [38]3 years ago
8 0
Video game designer for sure
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Going back the b beginning of the process is common in engineering true or false?
Yuliya22 [10]

Answer:

true !!

Explanation:

hope i helped :)! brainliest ?

6 0
3 years ago
Earth whose in situ weight is 105lb/cf and whose compacted weight is 122 lb/cf is placed in a fill at the rate of 260 cy/hr, mea
Stells [14]

Answer:

Number of rollers required to complete the compaction are 2

Explanation:

The solution is given in the attachments.

6 0
3 years ago
Showing or hiding records in a database is called “filtering.”<br> True<br> False
agasfer [191]

Answer:

TRUE

Explanation:

4 0
2 years ago
Read 2 more answers
A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 m
natima [27]

Answer:

a. 0.4544 N

b. 5.112 \times 10^{-5 M}

Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

The balanced reaction

H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O

NaOH\ Mass = Normality \times equivalent\ weight \times\ volume

= 0.3200 \times 40 g \times 21.30 mL \times  1L/1000mL

= 0.27264 g

NaOH\ mass = \frac{mass}{molecular\ weight}

= \frac{0.27264\ g}{40g/mol}

= 0.006816 mol

Now

Moles of H_2SO_4 needed  is

= \frac{0.006816}{2}

= 0.003408 mol

Mass\ of\ H_2SO_4 = moles \times molecular\ weight

= 0.003408\ mol \times 98g/mol

= 0.333984 g

Now based on the above calculation

a. Normality of acid is

= \frac{acid\ mass}{equivalent\ weight \times volume}

= \frac{0.333984 g}{49 \times 0.015}

= 0.4544 N

b. And, the acid solution molarity is

= \frac{moles}{Volume}

= \frac{0.003408 mol}{15\ mL \times  1L/1000\ mL}

= 0.00005112

=5.112 \times 10^{-5 M}

We simply applied the above formulas

4 0
3 years ago
Let be a real-valued signal for which when . Amplitude modulation is preformed to produce the signal . A proposed demodulation t
densk [106]

Answer:

hello your question is incomplete attached below is the complete question

answer : attached below

Explanation:

let ; x(t)  be a real value signal for x ( jw ) = 0 , |w| > 200\pi

g(t) = x ( t ) sin ( 2000 \pi t )

x_{1} (t) = \frac{1}{2}  x(t)  sin ( 4000\pi t )

next we apply Fourier transform

attached below is the remaining part of the solution

6 0
3 years ago
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