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3 years ago

Who invented the no zero

Mathematics

2 answers:

tiny-mole [99]3 years ago

8 0

Janna behanihe invented it

Vera_Pavlovna [14]3 years ago

7 0

Zero was invented independently by the Babylonians.

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Write 0.973 in Scientific Notation

yKpoI14uk [10]

Answer:

9.73 × 10-1

Step-by-step explanation:

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4 years ago

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Which graph shows the solution of the inequality −|x − 2| + 9 > 6?

mariarad [96]

answer is -1 < x < 5

-|x - 2|+ 9 > 6

Rearrange the terms

-|x - 2| > 6 - 9

-|x - 2| > - 3

then divide both sides of the inequality by the co- efficient of variable

|x - 2| < 3

convert the absolute inequality to standard inequality

-3 <x - 2 < 3

separate compound inequalities into system of inequality

{x - 2}> -3

{x - 2 < 3}

Rearrange variable to the left side of the equation

x > -3 + 2

calculate the sum or difference

x > -1

x -2 < 3

Rearrange variable to the left side of the equation

x < 3 + 2

calculate the sum or difference

x < 5

x > -1 and x < 5

Find intersection

-1 < x < 5

Equations and inequalities are both mathematical sentences formed by relating two expressions to each other. In an equation, the two expressions are deemed equal which is shown by the symbol =. Where as in an inequality, the two expressions are not necessarily equal which is indicated by the symbols: >, <, ≤ or ≥.

learn more about inequality equation here :https://brainly.in/question/15934172

SPJ9

4 0

1 year ago

Find the mean, variance, and standard deviation. If necessary, round your answer to the nearest hundredth.

-Dominant- [34]

Answer:

Mean:4

Variance:6.2

Standard deviation:2.5

5 0

3 years ago

Radical functions. please help me!

Sphinxa [80]

These are 8 questions and 8 answers:

1) Quesion 1:

9+√2
---------
4 - √7

Answer: the third option:

36 + 9√7 + 4√2 + √14
-----------------------------
9

Explanation:

Multiply both numerator and denominator by the conjugate of the denominator.

The conjugate of 4 - √7 = 4 + √7

=>

$\frac{9+ \sqrt{2} }{4- \sqrt{7} } . \frac{4+ \sqrt{7} }{4+ \sqrt{7} } = \frac{(9)(4)+9 \sqrt{7}+4 \sqrt{2} + \sqrt{2} . \sqrt{7} }{(4)^2-( \sqrt{7})^2 } =$

$= \frac{36+9 \sqrt{7} +4 \sqrt{2} + \sqrt{14} }{16-7}$

2) Question 2: sum

$5x (\sqrt[3]{x^2y})+2( \sqrt[3]{x^5y})$

Answer: fourth option

$7x( \sqrt[3]{x^2y} )$

Explanation:

Take x^5 out of the second radical which will result in a like term of the first radical:

$5x( \sqrt[3]{x^2y} )+2( \sqrt[3]{x^5y}) =5x( \sqrt[3]{x^2y} )+2x( \sqrt[3]{x^2y})=7x( \sqrt[3]{x^2y})$

which is the fourth option

3) Question 3. Which expression is equivalent to:

$\frac{ \sqrt{10} }{ \sqrt[4]{8} }$

Answer: the first option

Explanation

$\frac{ \sqrt{10} }{ \sqrt[4]{8} } = \frac{ \sqrt[4]{10^2} }{ \sqrt[4]{8} } = \frac{ \sqrt[4]{100} }{ \sqrt[4]{8} } . \frac{ \sqrt[4]{8^3} }{ \sqrt[4]{8^3} } = \frac{ \sqrt[4]{(100)(512)} }{8} = \frac{ \sqrt[4]{51200} }{8} = \frac{4 \sqrt[4]{200} }{8} = \frac{ \sqrt[4]{200} }{2}$

4) Question 4 What is the simplest form?

Answer: the second option

Explanation:

$\sqrt[4]{81x^8y^5}=x^2 y\sqrt[4]{3^4y} =3x^2y \sqrt[4]{y}$

5) Question 5 Product

Answer: the fourth option:

$104x^4+16x^4 \sqrt{30} [/tex]\\Explanation:\\Use the square of a binomial product: (a + b)^2 = a^2 + 2ab + b^2\\[tex](4x \sqrt{5x^2} )^2+2(4x \sqrt{5x^2})(2x^2 \sqrt{6}) +(2x^2 \sqrt{6} )^2=$

$=16x^2(5x^2)+16x^4( \sqrt{30} )+4x^4(6)=80x^4+16x^4 \sqrt{30} +24x^4=$

$=104x^4+16x^4 \sqrt{30}$

which is the fourth option.

6) Question 6 Product

Answer: fourth option

Explanation:

$\sqrt[3]{16x^7} . \sqrt[3]{12x^9} = \sqrt[3]{2^4.2^2.3x^7x^9} = \sqrt[3]{2^6.3.x^{16}}=2^2 x^5 \sqrt[3]{3x} =4 x^5\sqrt[3]{3x}$

which is the fourth option.

7) Question 7. Simplified form of 2√18 + 3√2 + √162

Answer: 18√2

Explanation:

$2 \sqrt{18}+3 \sqrt{2} + \sqrt{162}=2(3) \sqrt{2} + 3 \sqrt{2} +9 \sqrt{2} =18 \sqrt{2}$

which is the second option.

8) Question 8 which function is undefined for x = 0.

Answer: second option y = √ (x - 2)

Explanation.

The square root function is not defined for negative values.

When x = 0, x - 2 = -2, whose square root is not defined.

Therefore, the square root of x - 2 is not defined for x = 0.