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3 years ago

Drag each label to the correct location on the chart.

Mathematics

2 answers:

Over [174]3 years ago

8 0

Answer:

Step-by-step explanation:

boyakko [2]3 years ago

3 0

√(-5)^2 complex
400 real
-9+10i^2 real
0+5i complex
i^8 real
√-16 complex
-2+6i complex
√10 real

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1. Consider the inequality 2x < 6(2).

Arturiano [62]

We want

$2x$

So, if we divide both sides by 2, we have

$x$

So, we can accept all values for $x$ , as long as they are less than 6 (excluded)

6 0

3 years ago

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A boxcar contains six complex electronic systems. Two of the six are to be randomly selected for thorough testing and then class

zlopas [31]

Answer:

Step-by-step explanation:

Number of electronic systems = 6

(a) Number of defected systems = 2

Probability of getting at least one system is defective

1 defective and 1 non defective + 2 defective

= (2 C 1 ) x (4 C 1) + (2 C 2) / (6 C 2)

= 3 / 5

(b) four defective

Probability of getting at least one system is defective

2 defective and 2 non defective + 3 defective and 1 non defective + 4 defective

= (4 C 2 ) x (2 C 2) + (4 C 3 )(2 C 1) + (4 C 4) / (6 C 4)

= 1

4 0

3 years ago

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What is the following number in decimal form? 2.93 × 10−7 =

Arisa [49]

$a^{-n}=\left(\frac{1}{a}\right)^n\\\\10^{-7}=\left(\dfrac{1}{10}\right)^7=\dfrac{1}{10,000,000}=0.0000001\\\\2.93\times10^{-7}=2.93\times0.0000001=\boxed{0.000000293}$

3 0

4 years ago

So if I buy something for $26.98 and buy something else for $6.10 what will be the Total.

Ratling [72]

Answer:

33.08 $

I believe we add the two together and get our total !

— hope this helped :)!

6 0

3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago