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ser-zykov [4K]
3 years ago
6

What is (2+2+3+10)x(8+9+-9+7)

Mathematics
2 answers:
qwelly [4]3 years ago
6 0

Answer:

<em>255</em>

Step-by-step explanation:

(2+2+3+10) × (8+9+-9+7)

First, remove the brackets:

2 + 2 + 3 + 10 × 8 + 9 + -9 + 7

Now calculate like so:

2 + 2 + 3 + 10 = <em>17</em>

8 + 9 + -9 + 7 = <em>15</em>

<em>(</em><em>17</em><em>)</em><em> </em><em>×</em><em> </em><em>(</em><em>15</em><em>)</em><em> </em><em>=</em><em> </em><em>17</em><em> </em><em>×</em><em> </em><em>15</em><em> </em><em>=</em><em> </em><em>255</em>

<em>PLEASE</em><em> </em><em>DO</em><em> </em><em>MARK</em><em> </em><em>ME</em><em> </em><em>AS</em><em> </em><em>BRAINLIEST UWU</em><em> </em>

Norma-Jean [14]3 years ago
3 0

Answer:

Your answer for this question is 255.

Step-by-step explanation:

To solve this problem, we must remember how to use PEMDAS.  This tells us that we must simplify what is in parentheses first, exponents next, then multiplication and division, and finally addition and subtraction.  In this case, this means that we must first perform the operations inside of the parentheses before the multiplication of the two groups of parentheses.

If we simplify within both groups of parentheses, we get:

(2+2+3+10) * (8+9+-9+7)

= (17) * (15)

We get the above simplification by performing the addition of all of the constants in the first group of parentheses and performing the addition and subtraction in the second group (notice that the +9 and -9 cancel each other out).

Now, we must simply multiply together our final two values to obtain our final answer.

17 * 15 = 255

Therefore, your answer is 255.

Hope this helps!

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1) \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4} = \frac{1}{32}

3) (-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}

4) \frac{2}{2^{-4}} = 32

Step-by-step explanation:

1) \frac{(-2)^{-5}}{(-2)^{-10}}

Solving using exponent rule: a^{-m}=\frac{1}{a^m}

\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32

So, \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4}

Using the exponent rule: a^m.a^n=a^{m+n}

We have:

2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}

We also know that: a^{-m}=\frac{1}{a^m}

Using this rule:

2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}

So, 2^{-1}.2^{-4} = \frac{1}{32}

3) (-\frac{1}{2} )^3.(-\frac{1}{2} )^2

Solving:

(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}

So, (-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}

4) \frac{2}{2^{-4}}

We know that: a^{-m}=\frac{1}{a^m}

\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32

So, \frac{2}{2^{-4}} = 32

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