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dezoksy [38]
3 years ago
11

6300 and 530, 3 and 2 underlined Compare value The value 3 in____ is _____times

Mathematics
1 answer:
topjm [15]3 years ago
5 0
Basically when reading the value of a digit we start from the right.
Digits from the right to the left are named as follows:
units......>no zeros/digits after them
tens......> 1 zero/digit after the digit
hundreds.......> 2 zeros/digits after the digit
thousands......> 3 zeros/digits after the digit
ten thousands.......> 4 zeros/digits after the digit
hundred thousands.........> 5 zeros/digits after the digit
million........> 6 zeros/digits after the digit

Applying this on the above mentioned numbers, we will find that the 3 represents the hundreds (300) in 6300 while it represent the tens (30) in 530.

The question is to find the difference between the two, which means we will subtract them:
difference = 300 - 30 = 270

<span>The value 3 in 6300 is 270 times the value of 3 in 530
</span>
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Vinvika [58]

Answer:

8. m=6

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Step-by-step explanation:

Question 8:

2m-6=m

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<u>-6</u>=<u>-m</u>  

-       -     (divide both sides by the negative)

m=6

Question 9:

<u>8x</u>=<u>16  </u>              (divide both sides of the equation by 8)

8     8

x=2

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1 year ago
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sp2606 [1]

Answer:

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Step-by-step explanation:

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2 years ago
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Can we obtain a diagonal matrix by multiplying two non-diagonal matrices? give an example
polet [3.4K]
Yes, we can obtain a diagonal matrix by multiplying two non diagonal matrix.

Consider the matrix multiplication below

\left[\begin{array}{cc}a&b\\c&d\end{array}\right]   \left[\begin{array}{cc}e&f\\g&h\end{array}\right] =  \left[\begin{array}{cc}a e+b g&a f+b h\\c e+d g&c f+d h\end{array}\right]

For the product to be a diagonal matrix,

a f + b h = 0 ⇒ a f = -b h
and c e + d g = 0 ⇒ c e = -d g

Consider the following sets of values

a=1, \ \ b=2, \ \ c=3, \ \ d = 4, \ \ e=\frac{1}{3}, \ \ f=-1, \ \ g=-\frac{1}{4}, \ \ h=\frac{1}{2}

The the matrix product becomes:

\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}\frac{1}{3}&-1\\-\frac{1}{4}&\frac{1}{2}\end{array}\right] = \left[\begin{array}{cc}\frac{1}{3}-\frac{1}{2}&-1+1\\1-1&-3+2\end{array}\right]= \left[\begin{array}{cc}-\frac{1}{6}&0\\0&-1\end{array}\right]

Thus, as can be seen we can obtain a diagonal matrix that is a product of non diagonal matrices.
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3 years ago
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frosja888 [35]

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