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3 years ago

F(x)=x+3 , G(x)=x^2-2 (F/G)(-1)-G(3)

1 answer:

7 0

$f(x)=x+3;\ g(x)=x^2-2\\\\\left(\dfrac{f}{g}\right)(-1)-g(3)=?\\\\\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}\\\\g(3)=3^2-2=9-2=7\\\\f(-1)=-1+3=2\\\\g(-1)=(-1)^2-2=1-2=-1\\\\\left(\dfrac{f}{g}\right)(-1)-g(3)=\dfrac{2}{-1}-7=-2-7=-9$

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What does supplementary mean

murzikaleks [220]

Any two numbers, normally degrees, such as 10° and 170°, that add up to equal 180°

3 0

3 years ago

Solve the equation 5=1/2 v-3 <br><br> v=?

olya-2409 [2.1K]

5=1/2v-3
5=1/2v
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6 0

3 years ago

In an arithmetic sequence, the nth term an is given by the formula an=a1+(n−1)d, where a1 is the first term and d is the commo

Dmitry_Shevchenko [17]

Answer:

$a_{10} = \frac{10}{65536}$

Step-by-step explanation:

The first step to solving this problem is verifying if this sequence is an arithmetic sequence or a geometric sequence.

This sequence is arithmetic if:

$a_{3} - a_{2} = a_{2} - a_{1}$

We have that:

$a_{3} = 40, a_{2} = 10, a_{3} = \frac{5}{2}$

$a_{3} - a_{2} = a_{2} - a_{1}$

$\frac{5}{2} - 10 = 10 - 40$

$\frac{-15}{2} \neq -30$

This is not an arithmetic sequence.

This sequence is geometric if:

$\frac{a_{3}}{a_{2}} = \frac{a_{2}}{a_{1}}$

$\frac{\frac{5}[2}}{10} = \frac{10}{40}$

$\frac{5}{20} = \frac{1}{4}$

$\frac{1}{4} = \frac{1}{4}$

This is a geometric sequence, in which:

The first term is 40, so $a_{1} = 40$

The common ratio is $\frac{1}{4}$ , so $r = \frac{1}{4}$ .

We have that:

$a_{n} = a_{1}*r^{n-1}$

The 10th term is $a_{10}$ . So:

$a_{10} = a_{1}*r^{9}$

$a_{10} = 40*(\frac{1}{4})^{9}$

$a_{10} = \frac{40}{262144}$

Simplifying by 4, we have:

$a_{10} = \frac{10}{65536}$

3 0

3 years ago

Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the co

jeka57 [31]

Answer:

<h2>A. The series CONVERGES</h2>

Step-by-step explanation:

If $\sum a_n$ is a series, for the series to converge/diverge according to ratio test, the following conditions must be met.

$\lim_{n \to \infty} |\frac{a_n_+_1}{a_n}| = \rho$

If $\rho$ < 1, the series converges absolutely

If $\rho > 1$ , the series diverges

If $\rho = 1$ , the test fails.

Given the series $\sum\left\ {\infty} \atop {1} \right \frac{n^2}{5^n}$

To test for convergence or divergence using ratio test, we will use the condition above.

$a_n = \frac{n^2}{5^n} \\a_n_+_1 = \frac{(n+1)^2}{5^{n+1}}$

$\frac{a_n_+_1}{a_n} = \frac{{\frac{(n+1)^2}{5^{n+1}}}}{\frac{n^2}{5^n} }\\\\ \frac{a_n_+_1}{a_n} = {{\frac{(n+1)^2}{5^{n+1}} * \frac{5^n}{n^2}\$

$\frac{a_n_+_1}{a_n} = {{\frac{(n^2+2n+1)}{5^n*5^1}} * \frac{5^n}{n^2}\\$

aₙ₊₁/aₙ =

$\lim_{n \to \infty} |\frac{ n^2+2n+1}{5n^2}| \\\\Dividing\ through\ by \ n^2\\\\\lim_{n \to \infty} |\frac{ n^2/n^2+2n/n^2+1/n^2}{5n^2/n^2}|\\\\\lim_{n \to \infty} |\frac{1+2/n+1/n^2}{5}|\\\\$

note that any constant dividing infinity is equal to zero

$|\frac{1+2/\infty+1/\infty^2}{5}|\\\\$

$\frac{1+0+0}{5}\\ = 1/5$

$\rho = 1/5$

Since The limit of the sequence given is less than 1, hence the series converges.

5 0

3 years ago

Austin keeps a right conical basin for the birds in his garden as represented in the diagram. The basin is 40 centimeters deep,

Lina20 [59]

The shortest distance between the tip of the cone and its rim exits 51.11cm.

<h3>What is the shortest distance between the tip of the cone and its rim?</h3>

If you draw a line along the middle of the cone, you'd finish up with two right triangles and the line even bisects the angle between the sloping sides. The shortest distance between the tip of the cone and its rim exists in the hypotenuse of a right triangle with one angle calculating 38.5°. So, utilizing trigonometry and allowing x as the measurement of the shortest distance between the tip of the cone and its rim.

Cos 38.5 = 40 / x

Solving the value of x, we get

Multiply both sides by x

$$\cos \left(38.5^{\circ}\right) x=\frac{40}{x} x$