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Afina-wow [57]
3 years ago
6

Which technique is most appropriate to use to solve each equation?

Mathematics
2 answers:
Ann [662]3 years ago
6 0

Answer: The answers are (C) and (B).

Step-by-step explanation:  We are given two quadratic equations and we are to select the appropriate technique out of the given three options which is best to solve these equations separqtely.

The first quadratic equation is

x^2+4x=15.

Since 'x' term and the constant term both are present, so completing the square method will be appropriate to solve this equation. The solution is as follows:

x^2+4x=15\\\\\Rightarrow x^2+4x+4=15+4\\\\\Rightarrow (x+2)^2=19\\\\\Rightarrow x+2=\pm \sqrt {19}\\\\\Rightarrow x=-2\pm \sqrt{19}.

The second equation is

2x(x-3)=0.

Since the 'x' term is present but there is no constant term, so the zero product property will work here. The solution is as follows:

2x(x-3)=0\\\\\Rightarrow 2x=0,~x-3=0\\\\\Rightarrow x=0,~~x=3.

Thus, the correct options are (C) and (B).

allochka39001 [22]3 years ago
4 0
<span>x² + 4x = 15 can be solved best using the completing the square method.

2x(x−3)=0 can be best solved using the zero product property.
</span>
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Find the value of w (u(2))
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<h3>Answer:  17</h3>

==================================================

Explanation:

We'll start things off by computing the inner function u(2)

Plug x = 2 into the u(x) function

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This tells us that w(u(2)) is the same as w(-3). I replaced u(2) with -3.

We'll plug x = -3 into the w(x) function

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Therefore, w(u(2)) = 17

------------------------

Here's a slightly different approach:

Let's find what w(u(x)) is in general

w(x) = 2x^2 - 1

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w(u(x)) = 2(-x-1)^2 - 1

Then we can plug in x = 2

w(u(x)) = 2(-x-1)^2 - 1

w(u(2)) = 2(-2-1)^2 - 1

w(u(2)) = 2(-3)^2 - 1

w(u(2)) = 2(9) - 1

w(u(2)) = 18 - 1

w(u(2)) = 17

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Which expression is equivalent to 4(3n-5)?<br>A.7n - 9<br>B.7n - 20<br>C.12n-5<br>D.12n-20​
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