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Alexeev081 [22]
3 years ago
5

For every problem-solving activity it's crucial that no less than five alternatives be considered.

Mathematics
2 answers:
melisa1 [442]3 years ago
7 0

Problem-solving activity includes

1.Understanding the problem,that is nature of the problem, then Completely define in your own way.

2. Determining why this problem has accrued,

3.  Identifying the ways to solve the problem,

4. Prioritizing the given alternatives that is ways  and then arranging the alternatives for a solution,

5. Then applying the best solution or arrangement for the given problem.

There are two ways considered for  problem-solving activity

(1). Trial and Error  (2) Reduction in steps

It totally depends on the kind of problem , which you are solving. There may be Less than  five alternatives  ,equal to  five alternatives or more than  five alternatives to solve the problem.

Option B: False

matrenka [14]3 years ago
4 0
Yes this is true. For every problem-solving activity it's crucial that no less than five alternatives be considered.
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PLS ANSWER FAST I WILL MARK THE BEST ONE BRAINLIEST PLS
Fudgin [204]

Answer:

E

Step-by-step explanation:

if we decidedly factor this expression-

4x^2+4xy+y^2

= (2x)^2 + 2*2xy + y^2

= (2x + y)^2

as you can see it is a perfect square so it can be grouped into (2x+y)^2

hence your solution is option e

8 0
3 years ago
Read 2 more answers
Help! How would I solve this trig identity?
NeTakaya

Using simpler trigonometric identities, the given identity was proven below.

<h3>How to solve the trigonometric identity?</h3>

Remember that:

sec(x) = \frac{1}{cos(x)} \\\\tan(x) = \frac{sin(x)}{cos(x)}

Then the identity can be rewritten as:

sec^4(x) - sen^2(x) = tan^4(x) + tan^2(x)\\\\\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)}  = \frac{sin^4(x)}{cos^4(x)}  + \frac{sin^2(x)}{cos^2(x)} \\\\

Now we can multiply both sides by cos⁴(x) to get:

\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)}  = \frac{sin^4(x)}{cos^4(x)}  + \frac{sin^2(x)}{cos^2(x)} \\\\\\\\cos^4(x)*(\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)}) = cos^4(x)*( \frac{sin^4(x)}{cos^4(x)}  + \frac{sin^2(x)}{cos^2(x)})\\\\1 - cos^2(x) = sin^4(x) + cos^2(x)*sin^2(x)\\\\1 - cos^2(x) = sin^2(x)*sin^2(x) + cos^2(x)*sin^2(x)

Now we can use the identity:

sin²(x) + cos²(x) = 1

1 - cos^2(x) = sin^2(x)*(sin^2(x) + cos^2(x)) = sin^2(x)\\\\1 = sin^2(x) + cos^2(x) = 1

Thus, the identity was proven.

If you want to learn more about trigonometric identities:

brainly.com/question/7331447

#SPJ1

7 0
2 years ago
A chess board is an 8 by 8 grid of squares. If you randomly choose 3 squares on the board, what is the probability that:
bixtya [17]
The answer is the letter B
3 0
3 years ago
Let f ( ) sin(arctan ) x x = . What is the range of f ?
nikklg [1K]
<span>So you have composed two functions,
</span><span>h(x)=sin(x) and g(x)=arctan(x)</span>
<span>→f=h∘g</span><span>
meaning
</span><span>f(x)=h(g(x))</span>
<span>g:R→<span>[<span>−1;1</span>]</span></span>
<span>h:R→[−<span>π2</span>;<span>π2</span>]</span><span>
And since
</span><span>[−1;1]∈R→f is defined ∀x∈R</span><span>

And since arctan(x) is strictly increasing and continuous in [-1;1] ,
</span><span>h(g(]−∞;∞[))=h([−1;1])=[arctan(−1);arctan(1)]</span><span>
Meaning
</span><span>f:R→[arctan(−1);arctan(1)]=[−<span>π4</span>;<span>π4</span>]</span><span>
so there's your domain</span>
5 0
3 years ago
Write log4 14 as a logarithm of base 3. <br><br> HELP!!!!!
Sonja [21]
It would be log3 14/log3 4
Hope this helps. (:
3 0
3 years ago
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