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3 years ago

Decimal representation of a rational number cannot be

Mathematics

2 answers:

ANEK [815]3 years ago

6 0

Answer:

it would be an irrational number then

mezya [45]3 years ago

6 0

I think is (a) terminating

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What is the square root of r64

Fiesta28 [93]

Hope this helps! ;)

7 0

4 years ago

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A baker is making walnut bread. The recipe uses 2/3 cups of walnuts for each bread recipe. The baker has 7 1/2 cups of walnuts.

Lady_Fox [76]

Answer:

11 breads

Step-by-step explanation:

Note that:

1 bread recipe = 1 bread

From the above question, we know that:

2/3 cups of walnut = 1 bread

7 1/2 cups of walnut = x breads

Cross Multiply

2/3 cups × x breads = 7 1/2 cups × 1 bread

x breads = 7 1/2 cups × 1 bread/2/3 cups

x breads = 7 1/2 ÷ 2/3

x breads = 15/2 ÷ 2/3

x breads = 15/2 × 3/2

x breads = 45/4

x breads = 11 1/4 breads

Maximum full recipe = 11 breads.

Therefore, the MAXIMUM numbers of full recipe of walnut bread the baker can make is 11 breads.

7 0

3 years ago

5. Evaluate the function f (x) = 4 • 7x for x = -1 and x = 2. Show your work.

mylen [45]

F(-1)= 4*7(-1) = 4*-7 = -28

F(2)= 4*7(2) = 4*14 = 56

5 0

3 years ago

Select all the correct answers.<br> In which pairs of matrices does AB = BA?

horsena [70]

In order to multiply a matrix by another matrix, we multiply the rows in the first matrix by the columns in the other matrix (How this is done is shown below)

To determine the pairs of matrices that AB=BA, we will determine AB and BA for each of the options below.

For the first option

$A= \left[\begin{array}{cc}1&0&-2&1&\end{array}\right]; B= \left[\begin{array}{cc}5&0&3&2&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times5)+(0\times3)&(1\times0)+(0\times 2)&(-2\times5)+(1\times3)&(-2\times0)+(1\times2)&\end{array}\right]\\AB= \left[\begin{array}{cc}5+0&0+0&-10+3&0+2&\end{array}\right]\\AB = \left[\begin{array}{cc}5&0&-7&2&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(5\times1)+(0\times-2)&(5\times0)+(0\times 1)&(3\times1)+(2\times-2)&(3\times0)+(1\times2)&\end{array}\right]\\BA= \left[\begin{array}{cc}5+0&0+0&3+-4&0+2&\end{array}\right]\\BA = \left[\begin{array}{cc}5&0&-1&2&\end{array}\right] \\$

∴ AB≠BA

For the second option

$A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right]; B= \left[\begin{array}{cc}3&0&6&-3&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times3)+(0\times6)&(1\times0)+(0\times -3)&(-1\times3)+(2\times6)&(-1\times0)+(2\times-3)&\end{array}\right]\\AB= \left[\begin{array}{cc}3+0&0+0&-3+12&0+-6&\end{array}\right]\\AB = \left[\begin{array}{cc}3&0&9&-6&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(3\times1)+(0\times-1)&(3\times0)+(0\times 2)&(6\times1)+(-3\times-1)&(6\times0)+(-3\times2)&\end{array}\right]\\BA= \left[\begin{array}{cc}3+0&0+0&6+3&0+-6&\end{array}\right]\\BA = \left[\begin{array}{cc}3&0&9&-6&\end{array}\right] \\$

Here AB = BA

For the third option

$A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right]; B= \left[\begin{array}{cc}5&0&3&2&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times5)+(0\times3)&(1\times0)+(0\times 2)&(-1\times5)+(2\times3)&(-1\times0)+(2\times2)&\end{array}\right]\\AB= \left[\begin{array}{cc}5+0&0+0&-5+6&0+4&\end{array}\right]\\AB = \left[\begin{array}{cc}5&0&1&4&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(5\times1)+(0\times-1)&(5\times0)+(0\times 2)&(3\times1)+(2\times-1)&(3\times0)+(2\times2)&\end{array}\right]\\BA= \left[\begin{array}{cc}5+0&0+0&3+-2&0+4&\end{array}\right]\\BA = \left[\begin{array}{cc}5&0&1&4&\end{array}\right] \\$

Here also, AB=BA

For the fourth option

$A= \left[\begin{array}{cc}1&0&-2&1&\end{array}\right]; B= \left[\begin{array}{cc}3&0&6&-3&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times3)+(0\times6)&(1\times0)+(0\times -3)&(-2\times3)+(1\times6)&(-2\times0)+(1\times-3)&\end{array}\right]\\AB= \left[\begin{array}{cc}3+0&0+0&-6+6&0+-3&\end{array}\right]\\AB = \left[\begin{array}{cc}3&0&0&-3&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(3\times1)+(0\times-2)&(3\times0)+(0\times 1)&(6\times1)+(-3\times-2)&(6\times0)+(-3\times1)&\end{array}\right]\\BA= \left[\begin{array}{cc}3+0&0+0&6+6&0+-3&\end{array}\right]\\BA = \left[\begin{array}{cc}3&0&12&-3&\end{array}\right] \\$

Here, AB≠BA

Hence, it is only in the second and third options that AB = BA

$A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right] B= \left[\begin{array}{cc}3&0&6&-3&\end{array}\right] \\$ and $A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right]B= \left[\begin{array}{cc}5&0&3&2&\end{array}\right] \\$

Learn more on matrices multiplication here: brainly.com/question/12755004

8 0

3 years ago

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A group of tourists orders a special meal at a restaurant. The cost of each person's meal is $38. Each person will pay a $7 tip.

Iteru [2.4K]

Answer: To write an equation

y= 7x+38

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3 years ago

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