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DaniilM [7]
2 years ago
10

Solve the math problem

Mathematics
1 answer:
Kamila [148]2 years ago
6 0
The answer is -34 k^8 z^12.
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Evaluate the expression.<br><br> - 4(2 - 5)2 + 82
Pavlova-9 [17]

Answer:

106

Step-by-step explanation:

Ok this question is a bit confusing but once you understand the concept you will be able to answer the question correctly

Always know that bracket is solved first once it is found in the question

So back to the question

-4(2-5)2+82

So let's get to work...

Bracket first

-4(-3)2+82

Multiply with the number at the front then followed by the one at the back

(+12)2+82

24+82

106

Therefore the final answer is 106

I believe you understand how we get the answer...just be careful when solving this kind of question... simple

7 0
3 years ago
If you can answer it i'll give you brainliest
Nonamiya [84]

Answer:

2.

Step-by-step explanation:

The answer is the seocnd one.

6 0
3 years ago
If 5^n x 5^3 = 5^6, then n is _____.<br><br> A) 18<br> B) 2<br> C) 3<br> D) 9
Eddi Din [679]
<h3>Answer:  C) 3</h3>

The rule we'll use is a^b*a^c = a^(b+c). So we add the exponents.

That means 5^n*5^3 = 5^(n+3)

So 5^n*5^3 = 5^6 turns into 5^(n+3) = 5^6

The bases are equal to 5, so the exponents be equal to one another.

n+3 = 6

n+3-3 = 6-3

n = 3

So 5^3*5^3 = 5^(3+3) = 5^6.

6 0
3 years ago
Read 2 more answers
Determine whether the vectors u and v are parallel, orthogonal, or neither. u = &lt;6, -2&gt;, v = &lt;8, 24&gt;
Leno4ka [110]
Because 6 * 8 + ( - 2 )* 24 = 48 - 48 = 0 , the vectora u and v are orthogonal ;
Two vectors are orthogonal  if their dot product is zero.<span />
5 0
3 years ago
Check if (1,6) is the solution to the system shown below
Andre45 [30]

Answer:

no

Step-by-step explanation:

To determine if (1, 6) is a solution, substitute x = 1, y = 6 into the left side of both equations and if the value obtained equals the right side then it is a solution.

10(1) - 6 = 10 - 6 = 4 ≠ - 6

- 10(1) + 5(6) = - 10 + 30 = 20 ≠ 30

Then (1, 6 ) is not a solution to the system

4 0
3 years ago
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