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N76 [4]
3 years ago
8

NO LINKS!!! Please help fill in the blanks. Part 7a.​

Mathematics
1 answer:
inysia [295]3 years ago
6 0
<h3>Answer: Check out the screenshots below.</h3>

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

Explanations:

Part 1

For the first box, we use the log rule that log(A)+log(B) = log(A*B)

Then in the second box, we'll convert to exponential form. The logs are assumed to be base 10.

The third box then factors and uses the zero product property.

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

Part 2

We check each solution generated in part 1.

Plugging x = 5/3 will lead to the same number on each side. Therefore, x = 5/3 is a true solution.

In contrast, plugging x = -2 leads to a false equation. Recall that the domain of y = log(x) is x > 0. This means we cannot replace x with negative numbers. The value x = -2 is extraneous.

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

Part 3

There's not much to explain here that isn't already done so on the screenshot below.

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A proportion looks like this:

a\div b = c\div d

It means that the quantity a,b,c and d are in the same proportion: the ratio between a and b is the same ratio between c and d.

They are particularly useful in situation when you have an example given, and you want to extrapolate the relationship for another couple.

For example, you can think of the following problem: "Five apples cost 2 dollars. How much will 8 apples cost?"

You can build the proportion

5\div 2 = 8\div x

And then solve it for x:

\dfrac{5}{2}=\dfrac{8}{x} \iff 5x=16 \iff x = \dfrac{16}{5} = 3.2

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3 years ago
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For waht values of x do the vectors -1,0,-1), (2,1,2), (1,1, x) form a basis for R3?
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The values of x for which the given vectors are basis for R³ is:

                        x\neq 1

<h2>Step-by-step explanation:</h2>

We know that for a set of vectors are linearly independent if the matrix formed by these set of vectors is non-singular i.e. the determinant of the matrix formed by these vectors is non-zero.

We are given three vectors as:

(-1,0,-1), (2,1,2), (1,1, x)

The matrix formed by these vectors is:

\left[\begin{array}{ccc}-1&2&1\\0&1&1\\-1&2&x\end{array}\right]

Now, the determinant of this matrix is:

\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-1(x-2)-2(1)+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+2-2+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+1

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-x+1\neq 0\\\\\\i.e.\\\\\\x\neq 1

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3 years ago
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