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3 years ago

Find an equivalent fraction with the lowest common denominator. 9/10+2/3

Mathematics

2 answers:

mafiozo [28]3 years ago

8 0

Answer: 1-17/30, or 47/30

Step-by-step explanation:

Find the common denominator. In this case, it’s 30. Multiply the numerators to fit the new denominator. (10•3=30, so 9•3=27, and 2•10=20.)

Your new fractions are 27/30 and 20/30. Now add the numerators.

27+20=47.

Your new fraction is 47/30.

Now check if you can simplify. Since 47 is a prime number, this fraction is in its simplest form. However, it is an improper fraction, so you can simplify it to 1-17/30.

SIZIF [17.4K]3 years ago

6 0

Answer:

Step-by-step explanation:

9/10+2/3

The LCM = 30

9/10+2/3

= {9(3) + 2(10)}/30

= (27 + 20)/30

= 47/30

= 1 17/30

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2 years ago

Read 2 more answers

An electronics company produces transistors, resistors, and computer chips. Each transistor requires 3 units of copper, 1 unit

padilas [110]

Answer:

An electronics company can be produce 350 transistors and 340 computer chips, they can´t produce resistors.

Step-by-step explanation:

1. We will name the variables for transistors, resistors and the computer chips.

a = Transistors

b= Resistors

c = Computer chips

2. We propose three linear equations, one for the copper, one for the zinc and one for the glass.

$\left \{ {{3a+3b+2c=1730} \atop {a+2b+c=690}}\atop {2a+b+2c=1380}} \right.$

3. We write the matrix form as Ax=d

$A=\left(\begin{array}{ccc}3&3&2\\1&2&1\\2&1&2\end{array}\right)$

$x=\left(\begin{array}{ccc}a\\b\\c\end{array}\right)$

$A=\left(\begin{array}{ccc}1730\\690\\1380\end{array}\right)$

With this formula the solution of x is $x=\frac{d}{A}$ or $x=A^{-1}d$

4. We will find the inverse matrix $A^{-1}$ using the formula:

$A^{-1} = \frac{1}{detA} (C_{A})^{T}$

a. det A

$det A=\left[\begin{array}{ccc}3&3&2\\1&2&1\\2&1&2\end{array}\right] =3*(4-1)-3*(2-2)+2*(1-4)=9-0-6=3$

b. $(C_{A})^{T}$

$C_{A}=\left(\begin{array}{ccc}4-1&.(2-2)&1-4\\-(6-2)&6-4&-(3-6)\\3-4&-(3-2)&6-3\end{array}\right)$

$C_{A}=\left(\begin{array}{ccc}3&.0&-3\\-4&2&3\\-1&-1&3\end{array}\right)$

$(C_{A}) ^T=\left(\begin{array}{ccc}3&0&-3\\-4&2&3\\-1&-1&3\end{array}\right)^T$

$(C_{A}) ^T=\left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)$

c. $A^{-1}$

$A^{-1}=\frac{1}{3} \left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)$

5. As $x=\frac{d}{A}$ or $x=A^{-1}d$ , the solution of x is:

$x=\frac{1}{3}\left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)\left(\begin{array}{ccc}1730\\690\\1380\end{array}\right)$

$x=\frac{1}{3}\left(\begin{array}{ccc}(3*1730)+(-4*690)+(-1*1380)\\(0*1730)+(2*690)+(-1*1380)\\(-3*1730)+(3*690)+(3*1380)\end{array}\right)$

$x=\frac{1}{3}\left(\begin{array}{ccc}1050\\0\\1020)\end{array}\right)$

$X=\left[\begin{array}{ccc}350\\0\\340\end{array}\right]$

Therefore:

a= 350 Transistors

b=0 Resistors

c=340 Computer chips

4 0

2 years ago