7/9 Z - 6/9 + 6 -5 / 9 Z

What is the answer to<br> -12x+7=-29<br> x=?

Marat540 [252]

Answer: hope this will help you

Step-by-step explanation:

Can you mark me as brainlist plz

6 0

2 years ago

Read 2 more answers

Which list is in order from least to greatest? 1.94 times 10 Superscript negative 5, 1.25 times 10 Superscript negative 2, 6 tim

sergij07 [2.7K]

Answer:

The options in the question was not properly arranged/ formatted. See the arranged options in the explanation section

The correct option is A.

1.94 times 10 Superscript negative 5, (1.9 x 10⁻⁵)

1.25 times 10 Superscript negative 2 (1.25 x 10⁻²),

6 times 10 Superscript 4 (6 x 10⁴),

8.1 times 10 Superscript 4 (8.1 x 10⁴)

Step-by-step explanation:

Which list is in order from least to greatest?

1.94 times 10 Superscript negative 5,

1.25 times 10 Superscript negative 2,

6 times 10 Superscript 4,

8.1 times 10 Superscript 4

1.25 times 10 Superscript negative 2,

1.94 times 10 Superscript negative 5,

6 times 10 Superscript 4,

8.1 times 10 Superscript 4

1.25 times 10 Superscript negative 2,

1.94 times 10 Superscript negative 5,

8.1 times 10 Superscript 4,

6 times 10 Superscript 4

1.94 times 10 Superscript negative 5,

1.25 times 10 Superscript negative 2,

8.1 times 10 Superscript 4,

6 times 10 Superscript 4

Which list is in order from least to greatest?

A.

1.94 times 10 Superscript negative 5, (1.9 x 10⁻⁵)

1.25 times 10 Superscript negative 2 (1.25 x 10⁻²),

6 times 10 Superscript 4 (6 x 10⁴),

8.1 times 10 Superscript 4 (8.1 x 10⁴)

B

1.25 times 10 Superscript negative 2 (1.25 x 10⁻²),

1.94 times 10 Superscript negative 5 (1.9 x 10⁻⁵),

6 times 10 Superscript 4 (6 x 10⁴),

8.1 times 10 Superscript 4 (8.1 x 10⁴)

C.

1.25 times 10 Superscript negative 2 (1.25 x 10⁻²),

1.94 times 10 Superscript negative 5 (1.9 x 10⁻⁵),

8.1 times 10 Superscript 4 (8.1 x 10⁴),

6 times 10 Superscript 4 (6 x 10⁴)

D.

1.94 times 10 Superscript negative 5 (1.9 x 10⁻⁵),

1.25 times 10 Superscript negative 2 (1.25 x 10⁻²),

8.1 times 10 Superscript 4 (8.1 x 10⁴),

6 times 10 Superscript 4 (6 x 10⁴)

5 0

3 years ago

Read 2 more answers

Assume a trucker changes his tire every 40,000 miles, and that he starts with a brand new set of tires, how many sets of tires w

anyanavicka [17]

40 sets of tires. I learned this by looking up your question because it is incomplete.

6 0

2 years ago

Pls help???????? need to have an explanation also

Ratling [72]

Answer:

I would say the answer is 40 cookies

Step-by-step explanation:

The reason I say this is because if you write out the amount of cookies she can make with 6 cups of flour, you would get 30/6. 30 and 6 are both divisible by 6, so if you divided both by 6 you will get 5/1. 1 times 8 is 8, so if you multiply 5 by 8 you will get 40 cookies.

Hope this helped!

7 0

3 years ago

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

7/9 Z - 6/9 + 6 -5 / 9 Z - 9 simplify​

7/9 Z - 6/9 + 6 -5 / 9 Z - 9 simplify