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3 years ago

Write each rational number as a fraction for me PLEASEEEEE! 10 points

Mathematics

1 answer:

Nookie1986 [14]3 years ago

7 0

Answer:

Please check step by step explanation

Step-by-step explanation:

Here, we want to write each as a fraction

We proceed as follows;

a. -3/8 = -3/8

b. -5 = -5/1

c -51/4 = -21/4

d. -1.6 = -16/10 = -8/5

e. 8.03 = 803/100

f. 0.4 = 4/10

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What is the value of x

Sedbober [7]

Answer:

X = 56

Step-by-step explanation:

Subtracting 169 degrees from a straight line (180 degrees) gives you 11 degrees. with this information you can add 11 and 114 to get 124. Then you can subtract that amount from 180 (the measure of any triangle) and get your answer which is 56. I hope this helped!

4 0

3 years ago

Read 2 more answers

Point B lies between points A and C on Line segment A C . Let x represent the length of segment AB in inches. The length of line

tresset_1 [31]

AB = x

BC = 3x

AC = 20 inches = AB + BC

Let's plug our values into the equation above.

20 inches = x + 3x

Combine like terms

20 inches = 4x

Divide both sides by 4

5 = x

AB = x = 5 inches

BC = 3x = 3(5) = 15 inches

5 0

3 years ago

Given the equation:

Alisiya [41]

Answer:

x = 1 and x = $\frac{-4}{5}$

Step-by-step explanation:

In this equation, a=5, b=-1, c=-4.

Plug these into the quadratic formula:

x = $\frac{1+\sqrt{-1^{2}-(-4)(5)(-4) } }{2(5)}$ and x = $\frac{1-\sqrt{-1^{2}-(-4)(5)(-4) } }{2(5)}$

Now, simplify the equations:

x = $\frac{1+\sqrt{81} }{10}$ = $\frac{1+9}{10}$ = $\frac{10}{10}$ = 1

and

x = $\frac{1-\sqrt{81} }{10}$ = $\frac{1-9}{10}$ = $\frac{-8}{10}$ = $\frac{-4}{5}$

5 0

4 years ago

Let an = –3an-1 + 10an-2 with initial conditions a1 = 29 and a2 = –47. a) Write the first 5 terms of the recurrence relation. b)

zlopas [31]

We can express the recurrence,

$\begin{cases}a_1=29\\a_2=-47\\a_n=-3a_{n-1}+10a_{n-2}7\text{for }n\ge3\end{cases}$

in matrix form as

$\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}$

By substitution,

$\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}\implies\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^2\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}$

and continuing in this way we would find that

$\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}\begin{bmatrix}a_2\\a_1\end{bmatrix}$

Diagonalizing the coefficient matrix gives us

$\begin{bmatrix}-3&10\\1&0\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}$

which makes taking the $(n-2)$ -th power trivial:

$\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}^{n-2}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}$

$\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}$

So we have

$\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}\begin{bmatrix}a_2\\a_1\end{bmatrix}$