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3 years ago

What is the answer

Mathematics

1 answer:

Igoryamba3 years ago

8 0

Answer:

∠B = 60°

Step-by-step explanation:

They are supplementary angles, which means they add up to 180°.

∠B + ∠A = 180

? + 120 = 180

? = 180 - 120

? = 60°

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Phoebe compared the cost of books in 9 different stores.

krek1111 [17]

You have to put the numbers in order and find the middle value so in this case it's-

2, 4, 4, 5, 6, 7, 8, 8, 8

The answer is $6

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3 years ago

Answer 10,11 and 12 for 100 points

geniusboy [140]

10) 7(2m+3)
11) 3(3r-1)
12) 2p(5q+4)

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3 years ago

Read 2 more answers

Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions

monitta

I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take $z=y'$ , so that $z'=y''$ and we're left with the ODE linear in $z$ :

$y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2$

Now suppose $y$ has a power series expansion

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Then the ODE can be written as

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0$

$\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0$

All the coefficients of the series vanish, and setting $x=0$ in the power series forms for $y$ and $y'$ tell us that $y(0)=a_0$ and $y'(0)=a_1$ , so we get the recurrence

$\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}$

We can solve explicitly for $a_n$ quite easily:

$a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}$

and so on. Continuing in this way we end up with

$a_n=\dfrac{a_1}{n!}$

so that the solution to the ODE is

$y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x$

We also require the solution to satisfy $y(0)=a_0$ , which we can do easily by adding and subtracting a constant as needed:

$y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}$

4 0

3 years ago

3. A bag contains 4 white, 3 blue, and 6 red marbles. A marble is drawn from the bag, replaced,

gladu [14]

Answer:

Hope this helps! If I could have brainliest I'm trying to rank up.

Step-by-step explanation:

First find the probability of each color being drawn.

We need to find the total number of marbles first.

4 + 3 + 6 = 13

Everything will be over 13.

Probability of a white marble being drawn --> 4/13

Probability of a blue marble being drawn --> 3/13

Probability of a red marble being drawn --> 6/13

Now to find the probabilities of more than one being drawn, we can multiply.

Part A:

6/13*6/13=36/169

Part B:

3/13*3/13=9/169

Part C:

6/13*6/13=18/169

Part D:

7/13*7/13=49/169

8 0

3 years ago

If f(x) = -5-4 and g(x) = -3 -2 find (f-g)(x)

asambeis [7]

<u>Answer:</u>

The correct answer option is D. $-5^x+3x-2$

<u>Step-by-step-explanation:</u>

We are given the following two functions:

$f(x)=-5^x-4$

and

$g(x)= -3x-2$

We are to find $(f-g)(x)$ so we will subtract the function g from function f, arrange the like terms together and combine them like shown below to get:

$(f-g)(x) =(-5^x-4)-(-3x-2)$

$(f-g)(x)=-5^x-4+3x+2$

$(f-g)(x)=-5^x+3x-2$

Therefore, $(f-g)(x)$ is equal to $-5^x+3x-2$ so the correct answer option is D. $-5^x+3x-2$ .

5 0

3 years ago

What is the answer ​

What is the answer