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3 years ago

Convert the following fraction to a decimal.

Mathematics

1 answer:

Nookie1986 [14]3 years ago

6 0

I think it’s C but I could be wrong but I don’t think I am

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The height of a candle, in centimeters, can be modeled by the linear equation h = 28 - 2t, where t represents the hours that the

Ostrovityanka [42]

The slope is -2. In context to this problem, the candle will burn away 2cm of its length every hour.

5 0

3 years ago

Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.

sveticcg [70]

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

$\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =I.

First, we have to identify the matrix I. As it was said, the matrix is the identiy matrix, which means

I = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

So, $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Isolating the X, we have

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ - $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Resolving:

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]$

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X= $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ ⁻¹· $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ · $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a= $\frac{2}{11}$ ; b= $\frac{7}{33}$ ; c= $\frac{-1}{11}$ and d= $\frac{-3}{11}$ . Substituting:

X= $\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]$ · $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Multiplying the matrices, we have

X= $\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right]$

6 0

3 years ago

The sum of 3 consecutive even integers is 18

larisa [96]

Answer:

4,6,8

Step-by-step explanation:

Let x = 1st even integer

x+2 =2nd even integer

x+4 =3nd even integer

x+ x+2 + x+4 = 18

Combine like terms

3x+6 = 18

Subtract 6 from each side

3x+6-6=18-6

3x= 12

Divide by 3

3x/3 =12/3

x = 4

x+2 = 6

x+4 = 8

5 0

3 years ago

Simplly the square root of 75z^18

notsponge [240]

Answer:5√(3)z^9

Step-by-step explanation:

Square root of 75z^18

√(75z^18)

√(75) x √(z^18)

√(25x3) x (z^18)^(1/2)

√(25) x √(3) x z^(18x1/2)

5 x √(3) x z^9

5√(3)z^9

8 0

3 years ago

4.07065 in words or written form?

MAVERICK [17]

Four and seven thousand sixty-five hundred thousandths, I believe

3 0

3 years ago

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