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3 years ago

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Mathematics

1 answer:

Rashid [163]3 years ago

6 0

Answer:

Step-by-step explanation:

Evaluate f(- 2), then substitute the value obtained into g(x), that is

f(- 2) = - 2 + 2 = 0, then

g(0) = 3(0)² + 1 = 0 + 1 = 1

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Triangle+ Triangle + Triangle = 30 Triangle + circle + circle = 20 Circle + Square + Square = 13 Triangle + circle x half square

Rasek [7]

Answer: 20
Explanation: triangle=10 , circle= 5 , square= 4 , half square = 2....... 5 x 2 = 10 plus 10 = 20:)

4 0

2 years ago

Friday night a pizza parlor old five large pizza and ome medium pizza the Pizza parlor made a total of

DanielleElmas [232]

Friday night, a pizza parlor sold 5 large pizzas and some medium pizzas. The pizza parlor made a total of $291. How many medium pizzas were sold? Medium pizzas cost $9 and large pizzas cost $15

The large pizza costs $15 and earns $75, and the medium pizzas cost $9 and it earns $216. The large pizzas were sold 5, meanwhile, the medium pizzas were sold 24.

<h3>Why the pizza parlor sold 24 medium pizzas?</h3>

The pizza parlor made $291 from the large and medium pizzas.

The pizza parlor sold 5 large pizzas with cost $15.

5 × $15 = $75

$291 - $75 = $216

The pizza parlor sold … medium pizzas with cost $9.

$216 ÷ $9 = 24

The pizza parlor sold 24 medium pizzas.

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1 year ago

Power Series Differential equation

KatRina [158]

The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for $y$

$\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0$

which indeed gives the recurrence you found,

$a_{n+3}=-\dfrac{n-3}{n+3}a_n$

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that $a_2=0$ , and substituting this into the recurrence, you find that $a_2=a_5=a_8=\cdots=a_{3k-1}=0$ for all $k\ge1$ .

Next, the linear term tells you that $6a_0+6a_3=0$ , or $a_3=a_0$ .

Now, if $a_0$ is the first term in the sequence, then by the recurrence you have

$a_3=a_0$
$a_6=-\dfrac{3-3}{3+3}a_3=0$
$a_9=-\dfrac{6-3}{6+3}a_6=0$

and so on, such that $a_{3k}=0$ for all $k\ge2$ .

Finally, the quadratic term gives $6a_1-12a_4=0$ , or $a_4=\dfrac12a_1$ . Then by the recurrence,

$a_4=\dfrac12a_1$
$a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1$
$a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1$
$a_{13}=-\dfrac{10-3}{10+3}a_{10}=\dfrac{(-1)^3}2\dfrac{7\times4}{13\times10\times7}a_1$

and so on, such that

$a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}$

for all $k\ge2$ .

Now, the solution was proposed to be

$y=\displaystyle\sum_{n\ge0}a_nx^n$

so the general solution would be

$y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots$
$y=a_0(1+x^3)+a_1\left(x+\dfrac12x^4-\dfrac1{14}x^7+\cdots\right)$
$y=a_0(1+x^3)+a_1\displaystyle\left(x+\sum_{n=2}^\infty\left(\prod_{i=1}^{n-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\right)x^{3n-2}\right)$