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2 years ago

Solve for x, 3x+21=8x*54

Mathematics

2 answers:

maxonik [38]2 years ago

8 0

Answer:

$x=\frac{7}{143}$

Step-by-step explanation:

Simplify the right side of the equation. $3x+21=432x$ . Subtract 21 from both sides of the equation. $3x=432x-21$ . Subtract 432x from both sides of the equation, $-429x=-21$ . The negative signs cancel out to get $429x=21$ . Divide both sides but 429 to get $x=\frac{21}{429}$ . Reduce to get $x=\frac{7}{143}$ .

navik [9.2K]2 years ago

6 0

Answer:

Exact form: 7/143

Decimal form: /0.048951

You might be interested in

A school cotains 357 boys and 323 girls if a student is chosen at random, what is the probability that it is a girl

solong [7]

Answer:

$P(G)=0.475$

Step-by-step explanation:

From the question we are told that:

Number of boys $N_b=357$

Number of girls $N_g=323$

Total Number of students

$N_t=N_b+N_g$

$N_t=680$

Generally the equation for Probability of Choosing a Girl is mathematically given by

$P(G)=\frac{N_g}{N_t}$

$P(G)=\frac{323}{680}$

$P(G)=0.475$

3 0

3 years ago

Please solve this 20×10=

Vadim26 [7]

Answer:

20 × 10 = 200

Step-by-step explanation:

Multiply 2 × 1 = 2 and then add the two zeros onto the end to get 200.

3 0

3 years ago

At the beginning of summer, Mark has $200 in savings and Ryan has $350. The brothers have mowing jobs set up for the summer. Bec

xz_007 [3.2K]

Answer:

30 jobs

Step-by-step explanation:

Mark

Initial amount - $200
Earning - $20*j
Saving is: S = 20j + 200

Ryan

Initial amount - $350
Earning - $15*j
Saving is: S = 15j + 350

If they get same saving, then the equations will be equal:

20j + 200 = 15j + 350
20j - 15j = 350 - 200
5j = 150
j = 150/5
j = 30

The answer is 30 mowing jobs

5 0

3 years ago

An analog signal with a bandwidth of 36MHz is sampled at a frequency (fs) of 36,000,000 samples per second during the ADC (analo

Elza [17]

Answer:

The signal would have experienced aliasing.

Step-by-step explanation:

Given that:

the bandwidth of the signal $f_m$ = 36MHz

= 36 × 10⁶ Hz

The sampling frequency $f_s$ = 36 × 10⁶ Hz

Suppose the sampling frequency is equivalent to the bandwidth of the signal, then aliasing will occur.

Therefore, according to the Nyquist criteria;

Nyquist criteria posit that if the sampling frequency is more above twice the maximum frequency to be sampled, a repeating waveform can be accurately reconstructed.

∴

By Nyquist criteria, for perfect reconstruction of an original signal, i.e. the received signal without aliasing effect;

Then,

$f_s \geq 2f_m$

∴

The signal would have experienced aliasing.

4 0

3 years ago

HELP ME PLZ

joja [24]

Ok I think it’s 26 if I’m wrong sorry

7 0

3 years ago