All categories

3 years ago

Evaluate the expresssion |–12|–|8|.

Mathematics

2 answers:

Ilya [14]3 years ago

7 0

Answer:

Step-by-step explanation:

The lines around the -12 and 8 is what we call absolute value. This means that whatever is in it, it will be positive. Knowing this the equation would be 12-8. This would give you 4.

ruslelena [56]3 years ago

5 0

This essentially means 12-8
|| means absolute value which is the distance from 0 the number is
Basically it’s the number but positive

So the answer is 4

You might be interested in

What is f(x) = –4x2 + 24x + 13 written in vertex form? f(x) = –4(x – 6)2 + 23 f(x) = –4(x – 6)2 + 7 f(x) = –4(x – 3)2 + 4 f(x) =

nika2105 [10]

We have that
f(x) = –4x² + 24x + 13

we know that

The vertex form for a parabola that opens up or down is:

f(x) = a(x - h)^2 + k

in the given equation, a=-4, therefore we add zero to the original equation in the form of 4h²−4h²
f(x) = –4x² + 24x + 4h²−4h² +13
Factor 4 out of the first 3 terms and group them
f(x) = –4*(x² -6x +h²) +4h² +13
We can find the value of h by setting the middle term equal to -2hx
−2hx=−6x
h=3 and 4h²=36
f(x) = –4*(x² -6x +9) +36 +13

we know that the term (x² -6x +9) is equals to------> (x-3)²
so
f(x) = –4*(x-3)² +49

the answer is
f(x) = –4*(x-3)² +49

4 0

3 years ago

you are rolling a 10-sided die with sides numbered 1-10. you keep rolling the die until you roll a prime number and then stop. l

evablogger [386]

The formula for E(x) would be SUM[ X(n)P(n)] from n = 1 to n = infinity until we get the prime number

In the given problem, a 10-sided die with 1-10 numbers is rolled until we get a prime number

Prime numbers between 1-10 = 2,3,5,7 (4 prime numbers)

Non - prime numbers = 1,4,6,8,9,10( 6 non-prime number)

Let X be the number of times the die is rolled

The probability of getting a prime =

$P(Prime) = \frac{4}{10}$

Now, the value of

E(X)=∑x.P(x) [ x-> {1, infinity}]

= P(1)X(1) +P(2)X(2)+P(3)X(3)+P(4)X(4)........+ P(n)X(n)

= 1. $\frac{4}{10}$ + 2. $\frac{6}{10}$ . $\frac{4}{10}$ + 3. $\frac{6}{10}$ . $\frac{6}{10}$ $\frac{4}{10}$ + 4. $\frac{6}{10}$ . $\frac{6}{10}$ . $\frac{6}{10}$ . $\frac{4}{10}$ .......

Hence, this is the value of E(x)

To learn more about Prime number, here

brainly.com/question/9315685

#SPJ4

4 0

2 years ago

I will give Brainliest!!

sergey [27]

The value of y in the triangle is 6°.

<h3>How to calculate the value of y in the triangle?</h3>

Firstly, it's important to note that a triangle has the total angles of 180°.

Based on the information, the triangle has interior angle measures of 11y, 10y + 10°, and 10y - 16º.

The value of y in the triangle will be illustrated thus:

11y + 10y + 10 + 10y - 16 = 180

31y - 6 = 180

31y = 180 + 6.

31y = 186

Divide.

y = 186 / 31

y = 6

The value of y is 6.

Learn more about triangles on:

brainly.com/question/17335144

#SPJ1

5 0

1 year ago

Select the basic integration formula you can use to find the indefinite integral.

kondaur [170]

Answer:

$\int \dfrac{du}{u}=\log|u|+C$

$\int u^n du=\dfrac{u^{n+1}}{n+1}+C$

$\int \dfrac{du}{u\sqrt{u^2-a^2}}=\dfrac{1}{a}\csc^{-1}(\dfrac{x}{a})+C$

Step-by-step explanation:

$\int \dfrac{du}{u}=\log|u|+C$ $[\because \int \dfrac{dx}{x}=\log |x|+C]$

$\int u^n du=\dfrac{u^{n+1}}{n+1}+C$ $[\because \int x^n dx=\dfrac{x^{n+1}}{n+1}+C]$

$\int \dfrac{du}{u\sqrt{u^2-a^2}}=\dfrac{1}{a}\csc^{-1}(\dfrac{x}{a})+C$ $[\because \dfrac{adx}{x\sqrt{x^2-a^2}}=\csc^{-1}(\dfrac{x}{a})+C]$

3 0

3 years ago

Please solve this (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