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3 years ago

The question is attached.

Mathematics

2 answers:

vladimir1956 [14]3 years ago

7 0

Inverse inverartion

GuDViN [60]3 years ago

5 0

Answer:

inverse variation

Step-by-step explanation:

the numbers equal but then have to be devided

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A rental truck company charges a flat fee of $30 to rent a truck in addition to $0.25 per miles how much would it cost to rent a

Sidana [21]

Answer:

$50

Step-by-step explanation:

x is distance and y is cost because the cost is dependent on the distance which makes the distance the independent variable. you can always use y=mx+b for these problems!!

7 0

2 years ago

The population of a local species of bees can be found using an infinite geometric series where a1=860 and the common ratio is 1

erma4kov [3.2K]

$a_1=860$
$a_2=\dfrac{a_1}5$
$a_3=\dfrac{a_2}5=\dfrac{a_1}{5^2}$
$\vdots$
$a_n=\dfrac{a_{n-1}}5=\dfrac{a_{n-2}}{5^2}=\cdots=\dfrac{a_1}{5^{n-1}}$

The $k$ th partial sum is

$\displaystyle S_k=\sum_{n=1}^ka_n=\sum_{n=1}^k\frac{a_1}{5^{n-1}}$
$S_k=a_1\left(1+\dfrac15+\dfrac1{5^2}+\cdots+\dfrac1{5^{k-2}}+\dfrac1{5^{k-1}}$
$\dfrac15S_k=a_1\left(\dfrac15+\dfrac1{5^2}+\dfrac1{5^3}+\cdots+\dfrac1{5^{k-1}}+\dfrac1{5^k}\right)$

$\implies S_k-\dfrac15S_k=a_1\left(1-\dfrac1{5^k}\right)$
$\dfrac45 S_k=860\left(1-\dfrac1{5^k}\right)$
$S_k=1075-\dfrac{1075}{5^k}$

As $k\to\infty$ , we're left with

$\displaystyle\sum_{n=1}^\infty a_n=\lim_{k\to\infty}\left(1075-\frac{1075}{5^k}\right)=1075$

which is the upper limit to the population of the bees.

4 0

3 years ago

Identify the horizontal asymptote of f(x) =x2+5x-3/4x-1

Pachacha [2.7K]

since the numerator is x² + 5x - 3, and therefore has a degree of 2, whilst the denominator, 4x¹ - 1, has a degree of 1, therefore, there's no horizontal asymptote.

recall, we only get a horizontal asymptote if the denominator's expression degree is equals or greater than that of the numerator's.

5 0

4 years ago

Any idea on this please help

qwelly [4]

I think it might be 72+0?

8 0

4 years ago

Read 2 more answers

Sin α = 21/29, α lies in quadrant II, and cos β = 15/17, β lies in quadrant I Find sin (α - β).

Sever21 [200]

$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$

$\sin\alpha=\dfrac{21}{29}\implies \cos^2\alpha=1-\sin^2\alpha=\dfrac{400}{841}$

Since $\alpha$ lies in quadrant II, we have $\cos\alpha$ , so

$\cos\alpha=-\sqrt{\dfrac{400}{841}}=-\dfrac{20}{29}$

$\cos\beta=\dfrac{15}{17}\implies\sin^2\beta=1-\cos^2\beta=\dfrac{64}{289}$

$\beta$ lies in quadrant I, so $\sin\beta>0$ and

$\sin\beta=\sqrt{\dfrac{64}{289}}=\dfrac8{17}$

So

$\sin(\alpha-\beta)=\dfrac{21}{29}\dfrac{15}{17}-\left(-\dfrac{20}{29}\right)\dfrac8{17}=\dfrac{475}{493}$

8 0

3 years ago