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4 years ago

Use the graph to state the solution for the system.

Mathematics

1 answer:

FrozenT [24]4 years ago

3 0

We watch where the two lines intersect.

And that is at x = 3, and y = 0.

So it is (3, 0). Option C.

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What is the best estimate for the percent equivalent of 7/15

Sloan [31]

It is 46.6666667% 7/15=0.466666667 put into a percent it's 46.6666667%

6 0

3 years ago

Read 2 more answers

Which function does not have a vertical asymptote? A) y=(x) /(1-x²) . B) y=(5x) /(1-2x²) . C) y=(5x-1) /(3+x^2) . D) (5x) /(x+x²

uysha [10]

A function has a vertical asymptote $x=a$ at point a, where the denominator becomes equal to 0.

A. The denominator of the function $f(x)=\dfrac{x}{1-x^2}$ turns into 0 at $x=1$ or $x=-1.$ Then $x=1$ and $x=-1$ are two vertical asymptotes of this function.

B. The denominator of the function $f(x)=\dfrac{5x}{1-2x^2}$ turns into 0 at $x=\sqrt{\frac{1}{2}}$ or $x=-\sqrt{\frac{1}{2}}$ Then $x=\sqrt{\frac{1}{2}}$ and $x=-\sqrt{\frac{1}{2}}$ are two vertical asymptotes of this function.

C. The denominator of the function $f(x)=\dfrac{5x-1}{3+x^2}$ never turns into 0, then this function hasn't any asymptotes.

D. The denominator of the function $f(x)=\dfrac{x}{x+x^2}$ turns into 0 at $x=0.$ Then $x=0$ is vertical asymptote of this function.

Answer: correct choice is C.

5 0

3 years ago

If other factors are held constant, which combination of sample characteristics would produce the narrowest confidence interval

anzhelika [568]

Answer:

The correct option is C) large sample size (n) with small variance.

Step-by-step explanation:

Consider the provided information.

It is given that the other factors are held constant, and we want the narrowest confidence interval for a population mean.

Confidence interval for a population mean is directly proportional to variance and inversely proportional to the sample size.

If we increase the variance, CI will increase. But we want the narrowest CI, so variance should be small.

As CI is inversely proportional to sample size, therefore if we increase the sample size CI will decrease.

Hence, the correct option is C) large sample size (n) with small variance.

3 0

3 years ago

4.7x + 15.3x + 10.4 - 8.4

katrin2010 [14]

Answer:

10⋅(10x+1)

Hope this helps!

7 0

3 years ago

Lim n-> infinity [1/3 + 1/3² + 1/3³ + . . . .+ 1/3ⁿ]

Verizon [17]

Answer:

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]$

Let we first evaluate

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

Its a Geometric progression with

$\rm :\longmapsto\:a = \dfrac{1}{3}$

$\rm :\longmapsto\:r = \dfrac{1}{3}$

$\rm :\longmapsto\:n = n$

So, Sum of n terms of GP series is

$\rm :\longmapsto\:S_n = \dfrac{a(1 - {r}^{n} )}{1 - r}$

$\rm :\longmapsto\:S_n = \dfrac{1}{3} \bigg[\dfrac{1 - {\bigg[\dfrac{1}{3} \bigg]}^{n} }{1 - \dfrac{1}{3} } \bigg]$

$\rm :\longmapsto\:S_n = \dfrac{1}{3} \bigg[\dfrac{1 - {\bigg[\dfrac{1}{3} \bigg]}^{n} }{\dfrac{3 - 1}{3} } \bigg]$

$\rm :\longmapsto\:S_n = \dfrac{1}{3} \bigg[\dfrac{1 - {\bigg[\dfrac{1}{3} \bigg]}^{n} }{\dfrac{2}{3} } \bigg]$

$\bf\implies \:S_n = \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

<u>Hence, </u>

$\bf :\longmapsto\:\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } = \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

<u>Therefore, </u>

$\purple{\rm :\longmapsto\:\displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty }\rm \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

$\rm \: = \: \rm \dfrac{1}{2}\bigg[1 - 0 \bigg]$

$\rm \: = \: \rm \dfrac{1}{2}$

<u>Hence, </u>

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]} = \frac{1}{2}}}$

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<h3><u>Explore More</u></h3>

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1}}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1}}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1}}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1}}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga}}$

8 0

3 years ago