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3 years ago

Suppose y varies directly as x, and y = 8 when x = -2. Which of the following is the correct equation to set up to solve for the

constant of variation?
A) 8 = k(-2)

B) -2 = k(8)

C) -2/8 = k/1

D) k = (-2) - 8

Mathematics

1 answer:

slavikrds [6]3 years ago

7 0

$\bf \qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} y=8\\ x=-2 \end{cases}\implies 8=k(-2)$

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Erika is working on solving the exponential equation 50x = 17; however, she is not quite sure where to start. using complete sen

Kryger [21]

Erika is working on solving the exponential equation 50^x = 17; The solution is x = 0.7242.

<h3>What is logarithm?</h3>

When you raise a number with an exponent, there comes a result.

Let's say you get

a^b = c

Then, you can write 'b' in terms of 'a' and 'c' using a logarithm as follows

$b = \log_a(c)$

'a' is called base of this log function. We say that 'b' is the logarithm of 'c' to base 'a'

Erika is working on solving the exponential equation 50^x = 17;

however, she is not quite sure where to start.

The equation is

$50^x = 17\\$

By taking the log both sides

$log_a b = x\\\\a_x = b$

By using the base formula

$log_0 y = \dfrac{logy}{log b}$

So, $log_{50} 17= x\\\\$

$x = \dfrac{log17}{log 50}$

x = 0.7242

Learn more about logarithm here:

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2 years ago

DONT SEND A LINK TO DOWNLOAD FOR

Arte-miy333 [17]

Answer:

(6, -4) does not fit with these two equations

Step-by-step explanation:

y = 8 - 2x

x + 3y = 12

fit (6, -4) into the equations

-4 = 8 - 2(6)

8 - 2(6) = 8 - 12 = -4

works for this one

6 + 3(-4) = 12

6 + 3(-4) = 6 - 12 = -6

doesnt work for this one

8 0

3 years ago

What is the value of w?

solniwko [45]

If you have a no solution choice that will be your answer because w+34=w-22
you would subtract w from both sides to get 34=-22 and that's not true therefore it would be a no solution equation.

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3 years ago

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In ΔOPQ, the measure of ∠Q=90°, the measure of ∠O=84°, and OP = 2. 9 feet. Find the length of PQ to the nearest tenth of a foot

Afina-wow [57]

OP= 27. 6

Step-by-step explanatexplanation:

In triangle OPQ: Ô+P+Q =180

therefore <P= 6°

OQ= OP sinsinà

OP= OQ/ sinà

OP= 2.9/ sin6°

OP= 27.7 ft

using pythagoras

OP^2 = PQ^2 + OQ^2

PQ= sqrt( 27.74^2 - 2.9^2)

PQ= 27.27.6

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3 years ago

Lim n→∞[(n + n² + n³ + .... nⁿ)/(1ⁿ + 2ⁿ + 3ⁿ +....nⁿ)]

Schach [20]

Step-by-step explanation:

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

To, evaluate this limit, let we simplify numerator and denominator individually.

So, Consider Numerator

$\rm :\longmapsto\:n + {n}^{2} + {n}^{3} + - - - + {n}^{n}$

Clearly, if forms a Geometric progression with first term n and common ratio n respectively.

So, using Sum of n terms of GP, we get

$\rm \: = \: \dfrac{n( {n}^{n} - 1)}{n - 1}$

$\rm \: = \: \dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }$

Now, Consider Denominator, we have

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {n}^{n}$

can be rewritten as

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {(n - 1)}^{n} + {n}^{n}$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[{\dfrac{n - 1}{n}\bigg]}^{n} + \bigg[{\dfrac{n - 2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

Now, Consider

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

So, on substituting the values evaluated above, we get

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n} - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n}\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{1}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

Now, we know that,

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to \infty} \bigg[1 + \dfrac{k}{x} \bigg]^{x} = {e}^{k}}}}$

So, using this, we get

$\rm \: = \: \dfrac{1}{1 + {e}^{ - 1} + {e}^{ - 2} + - - - - \infty }$

Now, in denominator, its an infinite GP series with common ratio 1/e ( < 1 ) and first term 1, so using sum to infinite GP series, we have

$\rm \: = \: \dfrac{1}{\dfrac{1}{1 - \dfrac{1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{1}{ \dfrac{e - 1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{e}{e - 1} }$

$\rm \: = \: \dfrac{e - 1}{e}$

$\rm \: = \: 1 - \dfrac{1}{e}$

Hence,

$\boxed{\tt{ \displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} } = \frac{e - 1}{e} = 1 - \frac{1}{e}}}$

3 0

3 years ago