Answer:
1) (x + 3)(3x + 2)
2) x= +/-root6 - 1 by 5
Step-by-step explanation:
3x^2 + 11x + 6 = 0 (mid-term break)
using mid-term break
3x^2 + 9x + 2x + 6 = 0
factor out 3x from first pair and +2 from the second pair
3x(x + 3) + 2(x + 3)
factor out x+3
(x + 3)(3x + 2)
5x^2 + 2x = 1 (completing squares)
rearrange the equation
5x^2 + 2x - 1 = 0
divide both sides by 5 to cancel out the 5 of first term
5x^2/5 + 2x/5 - 1/5 = 0/5
x^2 + 2x/5 - 1/5 = 0
rearranging the equation to gain a+b=c form
x^2 + 2x/5 = 1/5
adding (1/5)^2 on both sides
x^2 + 2x/5 + (1/5)^2 = 1/5 + (1/5)^2
(x + 1/5)^2 = 1/5 + 1/25
(x + 1/5)^2 = 5 + 1 by 25
(x + 1/5)^2 = 6/25
taking square root on both sides
root(x + 1/5)^2 = +/- root(6/25)
x + 1/5 = +/- root6 /5
shifting 1/5 on the other side
x = +/- root6 /5 - 1/5
x = +/- root6 - 1 by 5
x = + root6 - 1 by 5 or x= - root6 - 1 by 5
The sound wave with a <u>frequency of 20</u> waves/sec is 800 longer than the wavelength of a sound wave with a <u>frequency of 16,000</u> waves/sec
<h3>Calculating wavelength </h3>
From the question, we are to determine how many times longer is the first sound wave compared to the second sound water
Using the formula,
v = fλ
∴ λ = v/f
Where v is the velocity
f is the frequency
and λ is the wavelength
For the first wave
f = 20 waves/sec
Then,
λ₁ = v/20
For the second wave
f = 16,000 waves/sec
λ₂ = v/16000
Then,
The factor by which the first sound wave is longer than the second sound wave is
λ₁/ λ₂ = (v/20) ÷( v/16000)
= (v/20) × 16000/v)
= 16000/20
= 800
Hence, the sound wave with a <u>frequency of 20</u> waves/sec is 800 longer than the wavelength of a sound wave with a <u>frequency of 16,000</u> waves/sec
Learn more on Calculating wavelength here: brainly.com/question/16396485
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Answer:
$40000
Step-by-step explanation:
640160≈640000
640000/16=40000
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Step-by-step explanation:
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