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3 years ago

Write the equation of the line that is perpendicular to the line 5y=x−5 through the point (-1,0).

Mathematics

1 answer:

Artemon [7]3 years ago

7 0

Hey there!

First, we want to put the equation of this first line in slope-intercept form.

5y=x-5

We divide both sides by 5.

y=1/5x-1.

The slope of a perpendicular is line is the negative reciprocal of the slope of the original line. The slope of a line perpendicular to a line with the equation y=1/2x would be -2, because you flip the numerator and denominator and then make it negative.

So, this means that the slope of the line perpendicular to the line y=1/5x-1 is -5. So, here's our equation so far.

y= -5x+b

Now, we need to find the b. To do this, we can plug in this point (-1,0) that this perpendicular line goes through and solve for b.

0=-5(-1)+b

0=5+b

b= -5

So, this gives us the equation y= -5x-5

Have a wonderful day!

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riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

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The difference of eighteen and a number is 3

Firdavs [7]

Answer:

Step-by-step explanation:

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18-3=x

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podryga [215]

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2 years ago

A survey of several 8 to 9 year olds recorded the following amounts spent on a trip to the mall: $31.11,$25.01,$18.53,$14.37,$24

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Solution :

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Confidence interval = 80%

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One sample T confidence interval

μ : Mean of variance

80% of confidence interval results :

Using statistical software,

Variable : data

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Std. Err. = 2.3479945

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Answer:

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Step-by-step explanation:

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