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Troyanec [42]
3 years ago
13

Find an equation for the line with the given properties. Express the equation in​ slope-intercept form.

Mathematics
1 answer:
Zanzabum3 years ago
4 0

Answer:

or, y + 4 = m (x - 7) + c is the given equation.

Step-by-step explanation:

Here, the slope of the equation = -5

⇒m = -5

The given point is (7, -4)

⇒(x_{0}, y_{0}) = (7,-4)

Now, the general form of the equation is :

y - y_{0}  = m(x - x_{0}) + c  : c = arbitrary constant

So, the given equation becomes

y - (-4)  = m(x - 7) + c

or, y+ 4 = m (x - 7) + c is the given equation.

Here, m = the slope of the equation.

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Step-by-step explanation:

Use Pythagorean Theorem to find the missing side

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a^{2} +b^{2} =c^{2}

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Does the series converge or diverge? If it converges, what is the sum? Show your work. ∑ ∞ n = 1 − 4 ( − 1 / 2 ) n − 1
Mekhanik [1.2K]

Answer:

Step-by-step explanation:

Given the series,

∑ ∞ n = 1 − 4 ( − 1 / 2 ) n − 1

I think the series is summation from n = 1 to ∞ of -4(-1/2)^(n-1)

So,

∑ − 4 ( − ½ )^(n − 1). From n = 1 to ∞

There are different types of test to show if a series converges or diverges

So, using Ratio test

Lim n → ∞ (a_n+1 / a_n)

Lim n → ∞ (-4(-1/ 2)^(n+1-1) / -4(-1/2)^(n-1))

Lim n → ∞ ((-4(-1/2)^(n) / -4(-1/2)^(n-1))

Lim n → ∞ (-1/2)ⁿ / (-1/2)^(n-1)

Lim n→ ∞ (-1/2)^(n-n+1)

Lim n→ ∞ (-1/2)^1 = -1/2

Since the limit is less than 0, then, the series converge...

Sum to infinity

Using geometric progression formula

S∞ = a / 1 - r

Where

a is first term

r is common ratio

So, first term is

a_1 = -4(-½)^1-1 = -4(-½)^0 = -4 × 1

a_1 = -4

Common ratio r = a_2 / a_1

a_2 = 4(-½)^2-1 = -4(-½)^1 = -4 × -½ = 2

a_2 = 2

Then,

r = a_2 / a_1 = 2 / -4 = -½

S∞ = -4 / 1--½

S∞ = -4 / 1 + ½

S∞ = -4 / 3/2 = -4 × 2 / 3

S∞ = -8 / 3 = -2⅔

The sum to infinity is -2.67 or -2⅔

Check attachment for better understanding

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What are the zeros of , where <br><br> ? help please need some help someone help please
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{ \boxed{\boxed{\begin{array}{cc} \maltese  \bf \: given \\  \\  \rm \: f(x) = ( {x}^{2}  + 16)( {x}^{2} - 9) \\  \\  \bf \: for \: zeroes \\  \\ \pink{ \boxed{\boxed{\begin{array}{c | c}   \bf \:  {x}^{2} + 16 = 0 & \bf \: {x}^{2}  - 9 = 0 \\  \\  =  >  {x}^{2}  =  - 16& {x}^{2}  = 9 \\  \\  =  > x =  \pm \sqrt{ - 16} &x =  \pm \:  \sqrt{9}  \\  \\  =  > x =  \pm \sqrt{ {i}^{2}  {4}^{2} } &x =  \pm \:  \sqrt{ {3}^{2} } \\  \\  =  > x =  \pm \: 4i&x =  \pm3  \end{array}}}}   \\  \\  \rm \: x =  \pm3 \: and \pm \: 4i\end{array}}}}

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