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4 years ago

If f(x) is a linear function, what is the value of n?

Mathematics

2 answers:

VikaD [51]4 years ago

7 0

Answer:

value of n is 5

Third option is correct.

Step-by-step explanation:

Let the equation of the linear function is y = mx + b

From the given table, we have two points (-4,-25) and (-1,-10)

The slope of the line containing these two points is

$m=\frac{y_2-y_1}{x_2-x_1}\\\\m=\frac{-10+25}{-1+4}\\\\m=\frac{15}{3}\\\\m=5$

Now, put m = 5, x = -4 and y = -25 in the equation y = mx+b

$-25=5\cdot(-4)+b\\\\-25=-20+b\\\\b = -5$

Therefore, equation of linear function is $y=5x-5$

Now, put x = n and y =20

$20=5n-5\\\\5n=25\\\\n=5$

Therefore, the value of n is 5

Viefleur [7K]4 years ago

6 0

m = (-10 + 25)/(-1 + 4) = 15/3 = 5

5 = (20 + 10)/(n + 1)

5(n+1) = 30

5n + 5 = 30

5n = 25

n = 5

Answer

n = 5

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4 years ago

A certain transverse wave is described by y(x,t)=bcos[2π(xl−tτ)], where b = 5.90 mm , l = 29.0 cm , and τ = 3.90×10−2 s .

Vitek1552 [10]

Part A:

The general form of the equation of a transverse wave is given by:

$y(x,t)=A\cos\left[2\pi\left( \frac{x}{\lambda} - \frac{t}{T} \right)\right]$

where A is the amplitude, $\lambda$ is the wavelength, and T is the period.

Given that a certain transverse wave is described by:

$y(x,t)=bcos[2\pi(xl-t\tau)]$ , where b = 5.90 mm , l = 29.0 cm , and \tau = 3.90\times10^{-2} s

Thus, the amplitude is b = 5.90 mm = 5.9\times10^{-3} \ m

Part B:

The general form of the equation of a transverse wave is given by:

$y(x,t)=A\cos\left[2\pi\left( \frac{x}{\lambda} - \frac{t}{T} \right)\right]$

where A is the amplitude, $\lambda$ is the wavelength, and T is the period.

Given that a certain transverse wave is described by:

$y(x,t)=bcos[2\pi\left(\frac{x}{l}-\frac{t}{tau}\right)\right]$ , where b = 5.90 mm , l = 29.0 cm , and \tau = 3.90\times10^{-2} s

Thus,

$y(x,t)=bcos[2\pi\left(\frac{x}{l}-\frac{t}{tau}\right)\right\\ \\ \frac{1}{\lambda} = \frac{1}{l} \\ \\ \Rightarrow\lambda= l =28.0 \ cm=\bold{2.8\times10^{-1}}$

Part C:

The general form of the equation of a transverse wave is given by:

$y(x,t)=A\cos\left[2\pi\left( \frac{x}{\lambda} - \frac{t}{T} \right)\right]$

where A is the amplitude, $\lambda$ is the wavelength, and T is the period.

Given that a certain transverse wave is described by:

$y(x,t)=bcos[2\pi\left(\frac{x}{l}-\frac{t}{tau}\right)\right]$ , where b = 5.90 mm , l = 29.0 cm , and \tau = 3.90\times10^{-2} s

The wave's frequency, f, is given by:

 $f= \frac{1}{T} = \frac{1}{\tau} = \frac{1}{3.40\times10^{-2}} =\bold{29.4 \ Hz}$

Part D:

Given that the the wavelength is $2.8\times10^{-1} \ m$ and that the wave's frequency is 29.4 Hz

The wave's speed of propagation, v, is given by:

$v=f\lambda=29.4(2.8\times10^{-1})=8.232 \ m/s$

4 0

3 years ago

Solve the system of equations. -3y + 5x = 26 – 2y – 5x =-16 X =? Y=?

Neko [114]

Answer:

x= 94/25

y= 58/5

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers