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1 year ago

Find the quotient: –3.5 and –0.875

Mathematics

2 answers:

romanna [79]1 year ago

8 0

Answer:

Step-by-step explanation:

-3.5/-0.875 = 4

vovangra [49]1 year ago

7 0

Answer:

The answer is 4 or C

Step-by-step explanation:

Since its a negative-negative equation, the result is a positive number. So take this equation as 3.5/0.875. Your result will be four or C. Hope this helps!

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1 year ago

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Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

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1 year ago

Show that b^2 is greater than equal to or less than ac, according as a, b, c are in A.P., G.P.

ValentinkaMS [17]

Answer/Step-by-step explanation (ac > b² or b² < ac. )

A/c to question, we have to show:-

b² >ac in A.P ........ (1)

b² = ac in G.P .....(2)

b² < ac in H.P. ..... (3)

b = a+c/2 (A.P)

b = √ac ( G.P)

b = 2ac/a+c (H.P)

In A.P :

b² > ac = b² - ac

= (a+c/2)² - ac

= (a²+2ac+c²/4) - ac = a² + 2ac + c² - 4ac / 4

= a² - 2ac + c² / 4 = ( a - c ) ² / 4 > 0 Hence, b²>ac

In G.P:-

b = √ac

Hence, b² = ac

In H.P :- b² < ac = ac > b² = ac - b² = ac - ( 2ac / a+c)

= ac(a+c) - 2ac / a+c

= a²c + ac² - 2ac / a+c

= ac(2ac - 2) / a+c > 0

Hence, ac > b² or b² < ac.

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1 year ago

4 and 5 are complements, and 1 and 5 are complements. By the congruent complements theorem, which angle is congruent to 4? 1 2 3

Zepler [3.9K]

The theorem tells you angles complementary to the same angle are congruent. Both 4 and 1 are complementary to 5, so 4 and 1 are congruent.

The angle congruent to 4 is 1.

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2 years ago

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In right triangle ABC, mC - 90° and AC BC. Which trigonometric ratio is cquivalent to sin b?

vaieri [72.5K]

Answer:

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