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4 years ago

HElp with this question PLEASE

Mathematics

2 answers:

maw [93]4 years ago

7 0

Answer:

14,899

Step-by-step explanation:

Find the value of t, then put that in the formula and do the arithmetic.

To find t, we need to find how many years 2039 is after 1988. This is done by subtraction:

... t = 2039 - 1988 = 51

Now, we put 51 into the formula where t is, then do the math.

... P(51) = 2500×e^(0.035×51)

... = 2500×e^1.785 . . . . . scientific calculators have a button for this

... ≈ 2500×5.95958

... ≈ 14,898.9 ≈ 14,899

_____

If your calculator doesn't do the exponential function, you can use a Google search box to do it.

AlladinOne [14]4 years ago

3 0

Answer: 14,899

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

P = 2500 e°⁰³⁵ˣ

t = 2039 - 1988

= 51

P = 2500 e°⁰³⁵⁽⁵¹⁾

= 2500 e¹°⁷⁸⁵

= 2500 (5.9596)

= 14,898.94

rounded to the nearest whole number: P = 14,899

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Determine if one of the given vectors is in the span of the other vectors.

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Answer:

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$\vec{b}=\begin{pmatrix}7\\ 3\\ 8\end{pmatrix}\\\vec{a}_{1}=\begin{pmatrix}1\\ 2\\ -4\end{pmatrix},\vec{a}_{2}=\begin{pmatrix}8\\ 5\\ 6\end{pmatrix},\vec{a}_{3}=\begin{pmatrix}3\\ 3\\ 0\end{pmatrix}$

Step-by-step explanation:

1) The point here is can this vector be written as a Linear Combination of the three other ones?

Writing it as matrices:

$\vec{b}=\begin{pmatrix}7\\ 3\\ 8\end{pmatrix}\\\vec{a}_{1}=\begin{pmatrix}1\\ 2\\ -4\end{pmatrix},\vec{a}_{2}=\begin{pmatrix}8\\ 5\\ 6\end{pmatrix},\vec{a}_{3}=\begin{pmatrix}3\\ 3\\ 0\end{pmatrix}$

Since we want to know if that can be a written as linear combination then let's rewrite them. This time, as an expression of vectors.

$\left\{\begin{matrix}x_{1}\vec{a}_{1}+x_{2}\vec{a}_{2}+x_{3}\vec{a}_{3}=\vec{b}\\ \end{matrix}\right.$

2) Or in other words can this equation be true?

$\begin{pmatrix}1 &8 &3 \\ 2 & 5 &3 \\ -4&6 &0 \end{pmatrix}=\begin{pmatrix}7\\ 3\\ 8\end{pmatrix}$

3) Augmenting that, and solving it by Gaussian Elimination:

$\left.\begin{pmatrix}1 &8 &3&|7 \\2 & 5& 3&|3\\ -4& 6 &0&|8 \\ \end{pmatrix}$

$2R_{1}-R_{2}\\-4R_{1}+R_{3}$

$R_{2}:-11\\8R_{2}\\R_{2}-R_{1}$

$38R_{2}-R_{3}$

(...) Then proceed the rest of the gaussian Eliminations, until you get the pivots on left side of the Matrix.

$\begin{pmatrix}1 &0 &0 &0 \\ 0 & 1 &0 & \frac{4}{3}\\ 0& 0 &1 & \frac{-11}{9}\end{pmatrix}$

$S=\left ( x_1=0,x_2= \frac{4}{3},x_3=\frac{-11}{9}\right )$

The solution are the scalar variables that may or may not be the span of those three other vectors as in this equation:

$\left\{\begin{matrix}x_{1}\vec{a}_{1}+x_{2}\vec{a}_{2}+x_{3}\vec{a}_{3}=\vec{b}\\ \end{matrix}\right.$

4) Finally, let's try it:

$0\begin{pmatrix}1\\ 2\\ -4\end{pmatrix}+\frac{4}{3}\begin{pmatrix}8\\ 5\\ 6\end{pmatrix}+\frac{-11}{9}\begin{pmatrix}3\\ 3\\ 0\end{pmatrix}=\begin{pmatrix}7\\ 3\\ 8\end{pmatrix}\\\\\begin{pmatrix}7\\ 3\\ 8\end{pmatrix}=\begin{pmatrix}7\\ 3\\ 8\end{pmatrix}$