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3 years ago

10

Which numbers are less than -8/3? Select all that apply.

1 answer:

AfilCa [17]3 years ago

6 0

Answer:

Step-by-step explanatiion:

-8/3=-2.67

Number less -2.7 are

-2.8; -2.9; -3.0; and so on

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Brainliest to whoever can answer questions correctly

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Answer:

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3 years ago

Match each number in scientific notation with the number written in standard form and in a step by step process.

geniusboy [140]

Answer:

1) Option 3

2) Option 1

3) Option 2

Step-by-step explanation:

To find : Match each number in scientific notation with the number written in standard form and in a step by step process.

Solution :

From scientific to standard we expand the power of 10.

1) $6.82\times 10^7$

Remove decimal and open the power,

$6.82\times 10^7=\frac{682}{100}\times 10000000$

$6.82\times 10^7=68200000$

Therefore, $6.82\times 10^7=68,200,000$

Match with Option 3.

2) $6.82\times 10^{-6}$

Remove decimal and open the power,

$6.82\times 10^{-6}=\frac{682}{100}\times \frac{1}{1000000}$

$6.82\times 10^{-6}=\frac{682}{100000000}$

Therefore, $6.82\times 10^{-6}=0.00000682$

Match with Option 1.

3) $6.82\times 10^6$

Remove decimal and open the power,

$6.82\times 10^6=\frac{682}{100}\times 1000000$

$6.82\times 10^6=6820000$

Therefore, $6.82\times 10^6=6,820,000$

Match with Option 2.

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3 years ago

The graph shows a bacteria population as a function of the number of days since an antibiotic was introduced.

miss Akunina [59]

14 million bacteria were present when the antibiotic was first introduced.

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3 years ago

Read 2 more answers

Problem 1: The Jefferson Band is selling baked goods to raise money for their trip. Their goal is to

Evgen [1.6K]

Answer:

334

Step-by-step explanation:

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3 years ago

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Which of the following are solutions to the equation below?

sladkih [1.3K]

Answer:

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=3,\:x=-\frac{1}{2}$

Step-by-step explanation:

considering the equation

$2x^2\:-\:4x\:-\:3\:=\:x$

solving

$2x^2\:-\:4x\:-\:3\:=\:x$

$2x^2-4x-3-x=x-x$

$2x^2-5x-3=0$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=-5,\:c=-3:\quad x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

solving

$x=\frac{-\left(-5\right)+\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$x=\frac{5+\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$x=\frac{5+\sqrt{49}}{2\cdot \:2}$

$x=\frac{5+7}{4}$

$x=3$

also solving

$x=\frac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$x=\frac{5-\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$x=\frac{5-\sqrt{49}}{4}$

$x=-\frac{2}{4}$

$x=-\frac{1}{2}$

Therefore,

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=3,\:x=-\frac{1}{2}$

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3 years ago