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3 years ago

11/6 = 1/3 + p (Find the common denominator first.)

Mathematics

1 answer:

vovikov84 [41]3 years ago

4 0

Answer:

p = 3/2

Step-by-step explanation:

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3 years ago

F(x)=x^2-9 how many roots in common

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3 years ago

Write the point-slope form of the line that passes through (6,1) and is parallel to a line with a slope of -3 include all of you

ycow [4]

<u>Answer: </u>

The point-slope form of the line that passes through (6,1) and is parallel to a line with a slope of -3 is 3x + y – 19 = 0

<u>Solution: </u>

The point slope form of the line that passes through the points $\left(x_{1} y_{1}\right)$ and parallel to the line with slope “m” is given as

$y - y_{1} = m\left(x - x_{1}\right)$ --- eqn 1

Where “m” is the slope of the line. $x_{1} \text { and } y_{1}$ are the points that passes through the line.

From question, given that slope “m” = -3

Given that the line passes through the points (6,1).Hence we get $x_{1} = 6 ; y_{1} = 1$

By substituting the values in eqn 1, we get the point slope form of the line which is parallel to the line having slope -3 can be found out.

y – 1 = -3(x – 6)

y – 1 = -3x +18

On rearranging the terms, we get

3x + y -1 – 18 = 0

3x + y – 19 = 0

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3 years ago

To find the equation of a line, we need the slope of the line and a point on the line. Since we are requested to find the equati

murzikaleks [220]

Answer:

$x-4y+4=0$

$f(x)=\sqrt x$ and x=4

Step-by-step explanation:

We are given that a curve

$y=\sqrt x$

We have to find the equation of tangent at point (4,2) on the given curve.

Let y=f(x)

Differentiate w.r.t x

$f'(x)=\frac{dy}{dx}=\frac{1}{2\sqrt x}$

By using the formula $\frac{d(\sqrt x)}{dx}=\frac{1}{2\sqrt x}$

Substitute x=4

Slope of tangent

$m=f'(x)=\frac{1}{2\sqrt 4}=\frac{1}{2\times 2}=\frac{1}{4}$

In given question

$m=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$

$\frac{1}{4}=\lim_{x\rightarrow 4}\frac{f(x)-f(4)}{x-4}$

By comparing we get a=4

Point-slope form

$y-y_1=m(x-x_1)$

Using the formula

The equation of tangent at point (4,2)

$y-2=\frac{1}{4}(x-4)$

$4y-8=x-4$

$x-4y-4+8=0$

$x-4y+4=0$

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