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Airida [17]
3 years ago
12

I need help on number 2

Mathematics
1 answer:
postnew [5]3 years ago
6 0

Answer:

Count the dots

Step-by-step explanation:

and you will get the nunber of kids

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How many solutions does the equation 3x − 7 = 4 + 6 + 4x have? (5 points) Two None Infinitely many One
statuscvo [17]

Answer:

There is one solution.

x = -17

<em>Proof and step-by-step explanation:</em>

Step 1: <em>Add the numbers</em>

3x-7 = 4+6+4x

3x-7 = 10+4x

Step 2: <em>Move</em><em> </em><em>terms</em>

3x-7 = 10+4x

3x-4x = 10+7

Step 3: <em>Collect the like terms and calculate the sum</em>

3x-4x = 10+7

-x = 17

Step 4: <em>Change the sign by multiplying both sides by -1</em>

-x (×-1) = 17 (×-1)

x = -17

I hope this helped ! :)

7 0
4 years ago
Write an expression for the sequence of operations described below. triple v, then subtract 7 from the result Do not simplify an
oee [108]

Given:

The sequence is defined as "triple v, then subtract 7 from the result".

To find:

The expression for the given sequence.

Solution:

Triple v means 3 times of v, i.e., 3v.

So, the result is 3v.

Then subtract 7 from the result. So, the expression for the sequence is

3v-7

Therefore, the required expression is 3v-7.

6 0
3 years ago
Factor 24x - 20. Fill in the missing numbers.<br><br> = 4( ___x - ____) .
Leya [2.2K]

Answer:

4(6x - 5)

Step-by-step explanation:

24/4 = 6

20/4 = 5

6 0
3 years ago
I need the answer fast pls
densk [106]
87 degrees is the answer
8 0
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
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