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Shalnov [3]
3 years ago
10

which of the following expressions is written in scientific notation? 0.3 • 104 2 • 10-5 10 • 103 12 • 10-3

Mathematics
1 answer:
netineya [11]3 years ago
8 0

Answer:

option (2) is correct.

2 \cdot 10^{-5} is in the scientific notation form.

Step-by-step explanation:

Given: Some expressions

1) \ 0.3 \cdot 10^4\\\\ 2) \ 2 \cdot 10^{-5}\\\\ 3) \ 10 \cdot 10^3\\\\ 4) \ 12 \cdot 10^{-3}\\\\

We have to choose which expressions is written in scientific notation.

For an expression to be in scientific notation the  most important requirement   is that the whole number is between 1 and 10, but not including 10.

that is  a\cdot 10^{n} , 1 ≤ a < 10 and n is integer.

Thus, out of given only 2 \cdot 10^{-5} is in the scientific notation form.

Thus, option (2) is correct.

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The coefficient of x is
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Answer:

coefficient is a number in front of a variable. For example, in the expression x 2-10x+25, the coefficient of the x 2 is 1 and the coefficient of the x is -10. The third term, 25, is called a constant.

3 0
2 years ago
Use emilimination to solve 2t 3n=9 and 5t-3n=5
Vladimir [108]
2t+3n=9                                                   2(2)+3n=9
+5t-3n=5 (the 3n cancel out)                   4+3n=9
7t=14                                                       3n=5
t=2                                                           n=5/3 or 1.67
3 0
3 years ago
A square matrix A is said to be idempotent iff A2 = A. (i) Show that if A is idempotent, then so is I − A. (ii) Show that if A i
muminat

Answer:

Step-by-step explanation:

Given that A is a square matrix and A is idempotent

A^2 = A

Consider I-A

i) (I-A)^2 = (I-A).(I-A)\\= I^2 -2A.I+A^2\\= I-2A+A\\=I-A

It follows that I-A is also idempotent

ii) Consider the matrix 2A-I

(2A-I).(2A-I)=\\4A^2-4AI+I^2\\= 4A-4A+I\\=I

So it follows that 2A-I matrix is its own inverse.

7 0
3 years ago
For which rational expression is 8 an excluded value ? check all that apply
denis23 [38]

<u>Answer:</u>

The correct answer options are C. \frac{x^2+5}{x-8} and D. \frac{x^2-x-56}{x^2-64}.

<u>Step-by-step explanation:</u>

The values which make the denominator equal to zero are called the excluded values.

Here, we can substitute 8 for x and check if it makes the denominator 0.

\frac{x-8}{x+8} = \frac{8-8}{8+8} =\frac{0}{16} =0

\frac{x-2}{x^2-4} = \frac{8-2}{8^2-4} =\frac{6}{60} =\frac{1}{10}

\frac{x^2+5}{x-8} = \frac{8^2+5}{8-8} =\frac{69}{0}

\frac{x^2-x-56}{x^2-64} = \frac{8^2-8-56}{8^2-64} = \frac{0}{0} =0

\frac{8x^2-2}{x^2-16} = \frac{8(8)^2-2}{8^2-16} =\frac{510}{48}

7 0
3 years ago
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The new price is $11.25
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