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2 years ago

The question is on the attached image :)

Mathematics

1 answer:

olga_2 [115]2 years ago

6 0

The mean can be found by adding the two numbers and dividing by 2

let the second number be y

(x + y)/2 = 1/2x + 1

solve for y

multiply each side by 2
x + y = 2(1/2x + 1)
distribute
x + y = x + 2
subtract x on both sides
y = 2

ANSWER: the second number is 2

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5, 3.1.24<br><br> f(x) = 3x² + 5; find f(x + 3) - [f(x) + 3]

Aleks [24]

Given that $f(x) = 3x^2 + 5$ , we have

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2 years ago

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3 years ago

Suppose you draw a card from a well shuffled deck of playing cards what is probability of drawing a ace

Tasya [4]

So there are a totally of 52 cards right?(not counting jokers because jokers are not in official cards)
And there are 4 aces right?
So there are 52 outcomes and 4 of them are aces.
So when we write probabilities we write it in fraction form where the total number of outcomes is the denominator and the ones you want to get are the numerator OK?
So the total number is outcomes is 52 and becomes the denominator and 4 the number of aces become the numerator.
So 4/52 right?
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3 years ago

Please solve this<br> (Rational numbers 8th grade)

pochemuha

Question # 1

Answer:

$\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=-\frac{3}{20}$

Step-by-step explanation:

Given the expression

$\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}$

$=-\frac{2}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}$ ∵ $\frac{-2}{3}\times \frac{3}{5}=-\frac{2}{5}$

$=-\frac{2}{5}+\frac{1}{4}$ ∵ $\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=\frac{1}{4}$

$\mathrm{Least\:Common\:Multiplier\:of\:}5,\:4:\quad 20$

$=-\frac{8}{20}+\frac{5}{20}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{-8+5}{20}$

$\mathrm{Add/Subtract\:the\:numbers:}\:-8+5=-3$

$=\frac{-3}{20}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=-\frac{3}{20}$

Therefore,

$\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=-\frac{3}{20}$

Question # 2

Answer:

$\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=-\frac{5}{28}$

Step-by-step explanation:

Given

$\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}$

$=-\frac{6}{35}-\frac{1}{6}\times \frac{3}{2}\times \frac{2}{5}\times \frac{1}{14}$ ∵ $\frac{2}{5}\times \frac{-3}{7}=-\frac{6}{35}$

$=-\frac{6}{35}-\frac{1}{140}$ ∵ $\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=\frac{1}{140}$

$\mathrm{Least\:Common\:Multiplier\:of\:}35,\:140:\quad 140$

$=-\frac{24}{140}-\frac{1}{140}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{-24-1}{140}$

$\mathrm{Subtract\:the\:numbers:}\:-24-1=-25$

$=\frac{-25}{140}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=-\frac{25}{140}$

$\mathrm{Cancel\:the\:common\:factor:}\:5$

$=-\frac{5}{28}$

Therefore,

$\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=-\frac{5}{28}$

4 0

3 years ago