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3 years ago

2+y-8=19 what is the value of y

1 answer:

8 0

The value of y would be 25. What you would do is add your 8 on both sides of the equal sign. After doing this your equation should look like, 2+y=27. Next you would subtract 2 on both sides of your equal sign and get 25. Your finishing equation should look like y=25.
Hope this helps! (:

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The solution of a system of linear equations is where the two lines intersect....

olchik [2.2K]

Answer:

True

Step-by-step explanation:

The place where lines intersect is where the x and y values match up.

Just like this, if the lines do not intersect, there's no solution.

If the lines seem to be one line, there is all real number solutions.

Hope this helped!

5 0

2 years ago

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Thomas sold 412 boxes of cookies at a cost of 1175 what is an estimate closest to the cost of a box of cookies

Nata [24]

Answer:

285

Step-by-step explanation:

1175 divided by 412 = 2.85194174757

that would be around 2.85

6 0

3 years ago

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Leroy invested 12,000 into two different bank accounts. One account (x) gives Leroy 1.5% interest yearly and the other account (

Anton [14]

0.015x+0.03(12000-x)=300
Solve for x
X=4000 at 1.5%

12000-4000=8000 at 3%

8 0

3 years ago

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Pls help to get the answer

soldier1979 [14.2K]

When they say quotient, they want a fraction, so the answer for that would be
$\frac{6}{10}$
and the decimal form of that would be 0.6. You can get that from dividing 6 by 10 and solving with long division.

3 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