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AlekseyPX
2 years ago
10

Find the product of 3 1/5 and 7 1/4. Express your answer in simplest form.

Mathematics
2 answers:
nata0808 [166]2 years ago
8 0
If I'm not mistaken, it should be 28/5 or 5 3/5 as the product is timesing the products together
ArbitrLikvidat [17]2 years ago
7 0
If you convert to an improper fraction, you will get 16/5 and 29/4. Since you are finding the product, which is multiplication, you would just multiply the numerator by the numerator (16x29) and the denominator by denominator (5x4) you will get 116/5 (simplified) . if you convert back, you will get 23 1/5
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3x−2y=−2 6x−2y=10 solve by elimination
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Answer:

x=4 y=7

Step-by-step explanation:

When multiplying 3x by -2, you get -6x. -6x and 6x cross out. You're left with -2y=-2 and -2y=10.

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A rectangle with a length of x+5 has a perimeter of 4x+ 14. What is the width of the rectangle?
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Translate the description as an algebraic expression:<br>the square of the ratio of 11 and k​
damaskus [11]

Answer:

The given description is equivalent to the algebraic expression,

(\frac {11}{k})^{2}

Step-by-step explanation:

The given description is ,

'The square of the ratio of 11 and k'

which is equivalent to the algebraic expression,

(\frac {11}{k})^{2}

6 0
3 years ago
Rewrite the system of linear equations as a matrix equation AX = B.
iren2701 [21]

Answer:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]

And the vector of vriables (X) is:

\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

In conclusion, AX=B is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

7 0
3 years ago
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