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1 year ago

The line x +ky=18 has a gradient of - 1/6 (a) Find the value of k.

Mathematics

2 answers:

Phoenix [80]1 year ago

7 0

Answer:

$k =6$

Step-by-step explanation:

$~~~~~~~x+ky = 18\\\\\implies ky = -x +18\\\\\implies y = -\dfrac 1k x+ \dfrac{18}k\\\\\text{It is now in the slope intercept form (y=mx +c), where m is the slope.}\\\\\text{Given that,}\\\\~~~~~m = -\dfrac 16\\\\\implies -\dfrac 1k = -\dfrac 16\\\\\implies \dfrac 1k = \dfrac 16\\\\\implies k =6$

Sophie [7]1 year ago

3 0

Answer:

k = 6

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope(gradient) and c the y- intercept )

given

x + ky = 18 ( subtract x from both sides )

ky = - x + 18 ( divide through by k )

y = - $\frac{1}{k}$ x + $\frac{18}{k}$ ← in slope- intercept form

with slope m = - $\frac{1}{k}$

given m = - $\frac{1}{6}$ , then equating

- $\frac{1}{k}$ = - $\frac{1}{6}$ ⇒ k = 6

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2 years ago

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3 years ago

Read 2 more answers

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Zero Slope

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2 years ago

Set up the integral that represents the arc length of the curve f(x) = ln(x) + 5 on [1, 3], and then use Simpson's Rule with n =

marta [7]

Answer:

The integral for the arc of length is:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx$

By using Simpon’s rule we get: 1.5355453

And using technology we get: 2.3020

The approximation is about 33% smaller than the exact result.

Explanation:

The formula for the length of arc of the function f(x) in the interval [a,b] is:

$\displaystyle\int_a^b \sqrt{1+[f'(x)]^2}dx$

We need the derivative of the function:

$f'(x)=\frac{1}{x}$

And we need it squared:

$[f'(x)]^2=\frac{1}{x^2}$

Then the integral is:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx$

Now, the Simposn’s rule with n=4 is:

$\displaystyle\int_a^b g(x)}dx\approx\frac{\Delta x}{3}\left( g(a)+4g(a+\Delta x)+2g(a+2\Delta x) +4g(a+3\Delta x)+g(b) \right)$

In this problem:

$a=1,b=3,n=4, \displaystyle\Delta x=\frac{b-a}{n}=\frac{2}{4}=\frac{1}{2},g(x)= \sqrt{1+\frac{1}{x^2}}$

So, the Simposn’s rule formula becomes:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\\\approx \frac{\frac{1}{3}}{3}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{1}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(1+\frac{2}{2}\right)^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{3}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)$

Then simplifying a bit:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx \approx \frac{1}{9}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(\frac{3}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(2\right)^2}} +4\sqrt{1+\frac{1}{\left(\frac{5}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)$

Then we just do those computations and we finally get the approximation via Simposn's rule:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\approx 1.5355453$

While when we do the integral by using technology we get: 2.3020.

The approximation with Simpon’s rule is close but about 33% smaller:

$\displaystyle\frac{2.3020-1.5355453}{2.3020}\cdot100\%\approx 33\%$

8 0

2 years ago

The line x +ky=18 has a gradient of - 1/6 (a) Find the value of k.​

The line x +ky=18 has a gradient of - 1/6 (a) Find the value of k.