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vladimir2022 [97]
3 years ago
15

Add. Express your answer in simplest form. 2/9 + 4/9

Mathematics
2 answers:
timama [110]3 years ago
5 0

2/9+4/9=6/9=2/3

Your answer is 2/3.

Vesnalui [34]3 years ago
3 0
The answer is 6/9 = 2/3
You might be interested in
You are going to roll two dice. Let the variable x = the sum of the numbers rolled. What is the probability that x = 9 ?
Vladimir79 [104]

The  first cube can land in any one of  6  ways.
The  second cube can land in any one of  6  ways.
Total number of ways that  2 dice can land = (6 x 6) = 36 ways.

For any of these ways, the sum of the numbers rolled is  9 :

3, 6
4, 5
5, 4
6, 3

There are  4  ways to roll a 9, out of a total of  36  ways that
the dice can land.  So the probability of rolling a  9  is

       4 / 36  =  1 / 9  =  <em>11-1/9 percent</em>


7 0
3 years ago
Can someone help me solve this? Pls and ty!
Zanzabum

Answer:

H. 95°

Step-by-step explanation:

m∠BGC is 10° smaller than m∠CGD, which means that m∠BGC is x° - 10°

m∠DGE is 25° more than m∠EGF, which means that m∠DGE is x° + 25°

The sum of all angles in a placement like this will always equal 360°

So we put all the measures of all angles on one side of thee equation and set it equal to 360*

x° + x° - 10° + 2x° + 25° + 2x° + 3x° + 30° = 360°

After combining like terms we get 9x° + 45° = 360°

After some algebra we get x = 35°

Now at the start we said that m∠DGE is x + 25°

Plug in 35° for x and we get 35° + 25°

The final answer is m∠DGE = 95°

Edit: Sorry if my answer looks a bit confusing. This is actually my first answer on Brainly so I'm quite new to this experience.

5 0
2 years ago
Read 2 more answers
If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.
Pie
Given that f^{(n)}(0)=(n+1)!, we have for f(x) the Taylor series expansion about 0 as

f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n

Replace n+1 with n, so that the series is equivalent to

f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}

and notice that

\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}

Recall that for |x|, we have

\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}

which means

f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}
\implies f(x)=\dfrac1{(1-x)^2}
5 0
3 years ago
Name
miv72 [106K]
Hggyuihdsryui um sorry but I dont no
8 0
3 years ago
Can someone help ? ASAP thanks
Ahat [919]

Answer:

2,5 then 17 20 then keep adding 15 for x and y

6 0
3 years ago
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