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MAXImum [283]
3 years ago
13

HELP ME WITH THIS PLEASE PLEASE SHOW ME THE FORMULA FOR LETTER C​

Mathematics
2 answers:
OLEGan [10]3 years ago
5 0

Answer:

Which subject is this . please tell

larisa86 [58]3 years ago
5 0

Answer:

see explanation

Step-by-step explanation:

1

(a)

Calculate slope m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁  = (8, 3) and (x₂, y₂ ) = (10, 7)

m = \frac{7-3}{10-8} = \frac{4}{2} = 2

(b)

Given a line with slope m then the slope of a line perpendicular to it is

m_{perpendicular} = - \frac{1}{m} = - \frac{1}{2}

(c)

The equation of a line in slope- intercept form is

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

Here m = - \frac{1}{2} , then

y = - \frac{1}{2} x + c ← is the partial equation

To find c substitute (8, 3) into the partial equation

3 = - 4 + c ⇒ c = 3 + 4 = 7

y = - \frac{1}{2} x + 7 ← equation of perpendicular line

--------------------------------------------------------------------------

2

(a)

with (x₁, y₁ ) = (3, 5) and (x₂, y₂ ) = (4, 4)

m = \frac{4-5}{4-3} = \frac{-1}{1} = - 1

(b)

m_{perpendicular} = - \frac{1}{-1} = 1

(c)

y = x + c ← is the partial equation

To find c substitute (3, 5) into the partial equation

5 = 3 + c ⇒ c = 5 - 3 = 2

y = x + 2 ← equation of perpendicular line

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maks197457 [2]

Answer:

x3squared-3x2squared-18x

Step-by-step explanation:

1 Expand by distributing terms.

({x}^{2}-6x)(x+3)(x

​2

​​ −6x)(x+3)

2 Use the FOIL method: (a+b)(c+d)=ac+ad+bc+bd(a+b)(c+d)=ac+ad+bc+bd.

{x}^{3}+3{x}^{2}-6{x}^{2}-18xx

​3

​​ +3x

​2

​​ −6x

​2

​​ −18x

3 Collect like terms.

{x}^{3}+(3{x}^{2}-6{x}^{2})-18xx

​3

​​ +(3x

​2

​​ −6x

​2

​​ )−18x

4 Simplify.

{x}^{3}-3{x}^{2}-18xx

​3

​​ −3x

​2

​​ −18x

5 0
3 years ago
Please help mee! The picture is up there :)
Gala2k [10]

Answer:0

the answer is 5

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Complete the steps to solve the inequality:<br><br> 0.2(x + 20) – 3 &gt; –7 – 6.2x
KIM [24]

Answer:

x > -1.25

Step-by-step explanation:

First, let's start with the left side of the equation.

1) multiply 0.2(x  + 20). You will get 0.2x+4

So you have 0.2x+4-3

Simplify that, you will have 0.2x+1

Now, we need to isolate the variable (bring all terms with "x" to one side), and move everything else to another side. Remember that when you bring something to the other side, you must change the sign in front of the term (for example, bringing 2x to another side would change it to -2x. another example is if you were to bring -2 to another side, you would have to change it to 2.)

2) 0.2x+6.2x>-7-1               Moved like terms to one side.                    

6.4x>-8                                 I combined the terms here!

x > -1.25                                 Simplified!

Let me know if you need anything else :)

5 0
3 years ago
Read 2 more answers
What are all the roots of the function f(x)= x^3+3x^2-x-3
frozen [14]

Answer:

\mathrm{Domain\:of\:}\:x^3+3x^2-x-3\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:<x<\infty \\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}

Step-by-step explanation:

4 0
2 years ago
D/d{cosec^-1(1+x²/2x)} is equal to​
SIZIF [17.4K]

Step-by-step explanation:

\large\underline{\sf{Solution-}}

\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

Let assume that

\rm :\longmapsto\:y =  {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

We know,

\boxed{\tt{  {cosec}^{ - 1}x =  {sin}^{ - 1}\bigg( \dfrac{1}{x} \bigg)}}

So, using this, we get

\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2x}{1 +  {x}^{2} } \bigg)

Now, we use Method of Substitution, So we substitute

\red{\rm :\longmapsto\:x = tanz \: \rm\implies \:z =  {tan}^{ - 1}x}

So, above expression can be rewritten as

\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2tanz}{1 +  {tan}^{2} z} \bigg)

\rm :\longmapsto\:y = sin^{ - 1} \bigg( sin2z \bigg)

\rm\implies \:y = 2z

\bf\implies \:y = 2 {tan}^{ - 1}x

So,

\bf\implies \: {cosec}^{ - 1}\bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg) = 2 {tan}^{ - 1}x

Thus,

\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

\rm \:  =  \: \dfrac{d}{dx}(2 {tan}^{ - 1}x)

\rm \:  =  \: 2 \: \dfrac{d}{dx}( {tan}^{ - 1}x)

\rm \:  =  \: 2 \times \dfrac{1}{1 +  {x}^{2} }

\rm \:  =  \: \dfrac{2}{1 +  {x}^{2} }

<u>Hence, </u>

\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg) =  \frac{2}{1 +  {x}^{2} }}}}

<u>Hence, Option (d) is </u><u>correct.</u>

6 0
2 years ago
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