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2 years ago

HELP PLEASE

Mathematics

1 answer:

ale4655 [162]2 years ago

5 0

Answer:

The possible approximate frequencies of the guitar string are 516.82 Hz and 508.82 Hz

Step-by-step explanation:

The given period of the tuning fork, T₁ = 1.95 × 10⁻³ s

The number of beats in 6.0 s = 24 beats

Therefore, the beat frequency, $f_{beat}$ = 4 beats per second = 4 Hz

Let f₁ represent the frequency of the tuning fork, and let f₂ represent the frequency of the guitar string, we have;

f₁ = 1/T₁ = 1/(1.95 × 10⁻³) ≈ 512.82 Hz

We have;

$f_{beat} = \left | f_2 - f_1 \right |$

$\pm f_{beat} = f_2 - f_1$

Therefore, we get;

$4 \, Hz= \left | f_2 - 512.82 \, Hz \right |$

f₂ - 512.82 = ± 4

∴ f₂ = 512.82 ± 4

The possible frequencies of the guitar string, f₂ ≈ 516.82 Hz or f₂ ≈ 508.82 Hz

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Rewrite the expression using distributive property : 30x + 6

Anit [1.1K]

Answer:

This cannot be done, but it can be factored as: 6(5x + 1)

Step-by-step explanation:

This expression cannot be distributed any further.

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5(x + 2) = 5x + 10

We can factor this expression however to go the opposite direction:

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Since 30 and 6 are both divisible by 6, we can factor it out of each term.

Divide 30x by 6 and divide 6 by 6.

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3 years ago

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Rainbow [258]

Answer:

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Step-by-step explanation:

Step1

Converting 431 to base 10.

4*102=400

3*101=30

1*100=1

Adding all to get Ans=43110

Step2 converting 43110 to 5

The equation calculation formula for 43110 number to 5 is like this below.

5|431

5|86|1

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Ans:32115

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3 years ago

Need help with inscribed angles If M

Bogdan [553]

Given:

In the given circle O, BC is diameter, OA is radius, DC is a chord parallel to chord BA and $m\angle BCD=30^\circ$ .

To find:

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If a transversal line intersect two parallel lines, then the alternate interior angles are congruent.

We have, DC is parallel to BA and BC is the transversal line.

$\angle OBA\cong \angle BCD$ [Alternate interior angles]

$m\angle OBA=m\angle BCD$

$m\angle OBA=30^\circ$

In triangle AOB, OA and OB are radii of the circle O. It means OA=OB and triangle AOB is an isosceles triangle.

We know that base angles of an isosceles triangle are congruent.

$\angle OAB\cong \angle OBA$ [Base angles of an isosceles triangle]

$m\angle OAB=m\angle OBA$

$m\angle OAB=30^\circ$

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$m\angle OAB+m\angle OBA+m\angle AOB=180^\circ$

$30^\circ+30^\circ+m\angle AOB=180^\circ$

$60^\circ+m\angle AOB=180^\circ$

$m\angle AOB=180^\circ-60^\circ$

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Therefore, the measure of angle AOB is 120 degrees.

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2 years ago

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3 years ago

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SashulF [63]

Answer:

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Add 3 °C to get a temperature 3 °C higher.

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3 years ago

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