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Sunny_sXe [5.5K]
2 years ago
11

If 3w +6=21 find the value of w

Mathematics
2 answers:
White raven [17]2 years ago
7 0

Answer:

3w + 6 = 21 \\ 3w = 15 \\ w = 5

Vikki [24]2 years ago
5 0

Hey there!

\mathsf{3w + 6 = 21}

<h3>SUBTRACT 6 to BOTH SIDES </h3>

\mathsf{3w + 6 - 6 = 21 - 6}

<h3>CANCEL out: 6 - 6 because that gives you 0 </h3><h3>KEEP: 21 - 6 because it helps you solve for w </h3>

\mathsf{21 - 6 = \boxed{\bf 15}}

<h3>NEW EQUATION: \mathsf{3w = 15}</h3><h3>DIVIDE 3 to BOTH SIDES </h3>

\mathsf{\dfrac{3w}{3}=\dfrac{15}{3}}

<h3>CANCEL out: \mathsf{\dfrac{3}{3}} because that gives you 1</h3><h3>KEEP: \mathsf{\dfrac{15}{3}} because that gives you the value of w</h3><h3>NEW EQUATION: \mathsf{\dfrac{15}{3}=w}</h3>

\mathsf{\dfrac{15}{3}=\boxed{\bf 5}}

\boxed{\boxed{\large\textsf{Therefore, your answer: \huge \bf w = 5}}}\huge\checkmark

\large\text{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

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Find all the zeros of the equation x^4-6x^2-7x-6=0 Explain please.
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<h3>Answer:</h3>
  • zeros are {-2, 3, (-1±i√3)/2}
<h3>Step-by-step explanation:</h3>

I like to look at a graph of the function to see where the zeros might be. Here, there are x-intercepts at x=-2 and x=3. These can be factored out using synthetic division to find the factorization to be ...

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By completing the square, using the quadratic formula, or by looking at the graph of it, the complex roots of the quadratic factor can be found to be ...

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The second attachment shows my synthetic division. The first division takes out the root x=3 to give a quotient of x³ +3x² +3x +2. The second division takes out the root -2 to give the quotient of x² +x +1. (You can see that I tried -1 as a root first.)

The graph shows both the quartic and the quadratic factor of it. The latter has a leading coefficient of 1 and a vertex at (-1/2, 3/4), so you know the complex roots are -1/2 ±i√(3/4).

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<em>From the beginning</em>

There is only a very complicated formula for the roots of a quartic equation, so these are usually solved by machine or by some form of trial and error (iteration). There are some helps, like Descarte's Rule of Signs, and the Rational Root theorem.

Here, the former looks at the one sign change in the coefficients to tell you there will be 1 positive real root. Changing the sign of the odd-degree terms makes there be 3 sign changes, so there will be 3 or 1 negative real roots. Thus, we're assured at least two real roots, one of each sign.

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<em>Comment on terminology</em>

"root" and "zero" are essentially the same thing when the function is equated to zero, as here. The terms refer to the value(s) of x that make the polynomial function evaluate to zero.

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