Answer:
(1,1) is a solution of the system.
Step-by-step explanation:
Let's solve the system.
y = 2x - 1
5x - 4y = 1
In the first equation, y is already separated as a function of x. So we replace in the second equation;
5x - 4(2x - 1) = 1
5x - 8x + 4 = 1
4 - 1 = 8x - 5x
x = 1
y = 2x - 1 = 2(1) - 1 = 1
(1,1) is a solution of the system.
Answer:
y = (-8/9)x + 0.77777
Step-by-step explanation:
We already know the slope, so the only thing left to find is the y-intercept.
To find the y-intercept, we can <u>plug in the slope and point to the slope-intercept form equation</u> (y = mx+b, where m=slope and b=y-int.)
y = m * x + b
(7) = (-8/9)*(-7) + b
<u>Now, just solve for b!</u>
(7) = (-8/9)*(-7) + b
7 = 56/9 + b
7 - (56/9) = b
b = 0.777777 (repeating decimal, usually signified by a little line above the 7)
so now we just <u>plug in the slope and y-intercept we found into y = mx + b.</u>
y = mx + b
y = (-8/9)x + 0.77777
Differentiating both sides of
with respect to <em>x</em> yields (using the chain rule)
Solve for d<em>y</em>/d<em>x</em> :
The answer is then D.
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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