Answer: I think it's C
Step-by-step explanation:
(83-64)-(64+29)
83-64=19
64+29=93
<u><em>19x-93</em></u>
Answer:
Simplifying
f(r) = 5 + 1.75r
Multiply f * r
fr = 5 + 1.75r
Solving
fr = 5 + 1.75r
Solving for variable 'f'.
Move all terms containing f to the left, all other terms to the right.
Divide each side by 'r'.
f = 5r-1 + 1.75
Simplifying
f = 5r-1 + 1.75
Reorder the terms:
f = 1.75 + 5r-1
Step-by-step explanation:
tada i think
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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Assuming that the cost per minute is the same for both months and the plan fee is the same, you can use y=mx+b for this
y is the cost of the phone plan, x is the cost per minute and b is the start cost.
so 19.41=25x+b for the first month
and 45.65=380x+b for the second month
solve both for b you get:
19.41-25x=b and 45.65-380x=b. from this we get
19.41-25x=45.65-380x
solve for x
328x=26.24 and x=0.08
this means the cost per minute is 0.08c/min (answer A)
rewrite the equation to calculate b, and where this time, the x is the number of minutes talked.
y=0.08x+b and plug in one of the two months
45.65=0.08 * 380 + b
Solve for b and b is 15.25
so the final equation is
y=0.08x+15.25 (answer B)
