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3 years ago

Work out m and c for the line: 3 y + 6 x = − 3

Mathematics

1 answer:

Usimov [2.4K]3 years ago

8 0

Answer:

m is -2 and c is -1

Step-by-step explanation:

• Let's first phrase out the general equation of a line

${ \rm{y = mx + c}}$

m is the slope
c is the y-intercept

[ remember that a general line equation must be in slope - intercept form as shown above ]

• from our question, we are given the equation;

${ \rm{3y + 6x = - 3}}$

• let's make y the subject in order to make the equation in slope - intercept format.

→ remember to apply "subject making knowledge"

${ \rm{3y = - 3 - 6x}} \\ \\ { \rm{3y = - 6x - 3}} \\ \\ { \rm{ \frac{3y}{3} = \frac{ - 6x}{3} - \frac{3}{3} }} \\ \\ { \boxed{ \rm{y = - 2x - 1}}}$

• The above boxed equation is now a general equation. Let's extract out slope, m and y-intercept, c

${ \rm{m \: \dashrightarrow \: - 2}} \\ \\ { \rm{c \: \dashrightarrow \: - 1}}$

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Some math help please?

Crank

Answer:

∠ LMP = 77° , ∠ NMP = 103°

Step-by-step explanation:

The 2 angles on the straight line sum to 180° , that is

∠ LMP + ∠ NMP = 180 , substitute values

- 16x + 13 - 20x + 23 = 180 , that is

- 36x + 36 = 180 ( subtract 36 from both sides )

- 36x = 144 ( divide both sides by - 36 )

x = - 4

Then

∠ LMP = - 16x + 13 = - 16(- 4) + 13 = 64 + 13 = 77°

∠ NMP = - 20x + 23 = - 20(- 4) + 23 = 80 + 23 = 103°

8 0

3 years ago

6. In the following right angled triangles, find the unknown sizes of the acute angles. Which no you can do plz do it .

Sedaia [141]

Answer:

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8 0

3 years ago

Use the definition of a Taylor series to find the first four nonzero terms of the series for f(x) centered at the given value of

Black_prince [1.1K]

Answer:

The first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ are:

$f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}$

Step-by-step explanation:

The Taylor series of the function f at a (or about a or centered at a) is given by

$f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$

To find the first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ you must:

In our case,

$f\left(x\right) \approx P\left(x\right) = \sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k=\sum\limits_{k=0}^{4}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$

So, what we need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.

$f^{(0)}\left(x\right)=f\left(x\right)=\frac{7}{x + 1}$

Evaluate the function at the point: $f\left(2\right)=\frac{7}{3}$

$f^{(1)}\left(x\right)=\left(f^{(0)}\left(x\right)\right)^{\prime}=\left(\frac{7}{x + 1}\right)^{\prime}=- \frac{7}{\left(x + 1\right)^{2}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime }=- \frac{7}{9}$

$f^{(2)}\left(x\right)=\left(f^{(1)}\left(x\right)\right)^{\prime}=\left(- \frac{7}{\left(x + 1\right)^{2}}\right)^{\prime}=\frac{14}{\left(x + 1\right)^{3}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime }=\frac{14}{27}$

$f^{(3)}\left(x\right)=\left(f^{(2)}\left(x\right)\right)^{\prime}=\left(\frac{14}{\left(x + 1\right)^{3}}\right)^{\prime}=- \frac{42}{\left(x + 1\right)^{4}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime \prime }=- \frac{14}{27}$

$f^{(4)}\left(x\right)=\left(f^{(3)}\left(x\right)\right)^{\prime}=\left(- \frac{42}{\left(x + 1\right)^{4}}\right)^{\prime}=\frac{168}{\left(x + 1\right)^{5}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime \prime \prime }=\frac{56}{81}$

Apply the Taylor series definition:

$f\left(x\right)\approx\frac{\frac{7}{3}}{0!}\left(x-\left(2\right)\right)^{0}+\frac{- \frac{7}{9}}{1!}\left(x-\left(2\right)\right)^{1}+\frac{\frac{14}{27}}{2!}\left(x-\left(2\right)\right)^{2}+\frac{- \frac{14}{27}}{3!}\left(x-\left(2\right)\right)^{3}+\frac{\frac{56}{81}}{4!}\left(x-\left(2\right)\right)^{4}$

The first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ are:

$f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}$

8 0

3 years ago

what base could be written in the blank to make the exponential function model 15% decay ? y= (__1/2__) ^t\12

Helga [31]

Answer:

0.85

Step-by-step explanation:

1 - 15% = 1 - 0.15 = 0.85

$y = (0.85)^\frac{t}{12}$

Answer: 0.85

5 0

1 year ago

The cost of performance tickets and beverages for a family of four can be modeled using the equation 4x + 12 = 48, where x repre

Kisachek [45]

4x + 12 = 48
4x = 48 -12
4x = 36
x= 36 / 4
x= 9

Each ticket is $9.

6 0

4 years ago