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anzhelika [568]
3 years ago
13

Brainly, 10+10=2 so what does 10+10+10= ?

Mathematics
2 answers:
maksim [4K]3 years ago
8 0
Banana wnwnskksksk77282
Vikki [24]3 years ago
4 0

Answer:

3

Step-by-step explanation:

1+0+1+0+1+0= 3

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Describe different ways that you can show a story problem
natima [27]
Once you know your basic operations addition<span>, </span>subtraction,multiplication<span>, </span>division <span>you will encounter story problems also known as word problems that require you to read a problem and decide which operation to perform in order to get the answer. There are key words here that often indicate which operation you will use. We will give you a list of them, but remember that for many problem</span><span>s </span><span>there may not be a key word and you’ll have to use your best judgment in order to figure out what to do. </span>
3 0
3 years ago
Whoever gets it right can have brainliest, but answer ASAP!
sveta [45]

Answer:

-29

Step-by-step explanation:

(-4 + -3) x 2^2 - 1

follow order of operations PEMDAS

Parentheses:

(-4 + -3) x 2^2 - 1

(-7) x 2^2 -1

exponents:

(-7) x 4 - 1

multiplication

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subtraction

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3 0
2 years ago
Read 2 more answers
Use the Distributive Property of Multiplication over Addition to find each missing number.
Svet_ta [14]

Answer:

(3*9)+(3*6)

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Best answer gets brainliest​
Luden [163]
K= -3

3y = x+6 can be rewritten as y = 1/3x + 2
a perpendicular slope is the opposite reciprocal of the original slope
so instead of 1/3 it would be -3
therefore. k must equal -3 to be perpendicular to 3y = x+6
3 0
3 years ago
Given that the series kcoskt kº +2 k=1 converges, suppose that the 3rd partial sum of the series is used to estimate the sum of
3241004551 [841]

Answer:

c

Step-by-step explanation:

Given that:

\sum \limits ^{\infty}_{k=1} \dfrac{kcos (k\pi)}{k^3+2}

since cos (kπ) = -1^k

Then, the  series can be expressed as:

\sum \limits ^{\infty}_{k=1} \dfrac{(-1)^kk)}{k^3+2}

In the sum of an alternating series, the best bound on the remainder for the approximation is related to its (n+1)^{th term.

∴

\sum \limits ^{\infty}_{k=1} \dfrac{(-1)^{(3+1)}(3+1))}{(3+1)^3+2}

\sum \limits ^{\infty}_{k=1} \dfrac{(-1)^{(4)}(4))}{(4)^3+2}

= \dfrac{4}{64+2}

=\dfrac{2}{33}

5 0
3 years ago
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