All categories

3 years ago

What do you do with fractions in finding absolute value

Mathematics

1 answer:

Furkat [3]3 years ago

4 0

Absolute value only removes the negative signs. Just multiply the numerator and the denominator

You might be interested in

Math, ,,,,,,,,,,,,,,,,,,,,

STatiana [176]

It’s 1/5 number a............

4 0

2 years ago

Read 2 more answers

Select the correct systems of equations.

jek_recluse [69]

(3 , -1)

Step by step explanation:

On the x-axis, the dot is below the number 3 and on the y-axis, the dot is on the right side of -1

8 0

2 years ago

Read 2 more answers

FIND PR.PLS HELP BRAINLIEST...

slega [8]

Answer:

6.2 units.

Step-by-step explanation:

We are given two angles and an opposite corresponding side. To solve for an unknown side, we can use the law of sines.

$\frac{8}{sin(90)}=\frac{x}{sin(180-39-90)}=\frac{x}{sin(51)}$

Cross multiplying and solving will get you this:

$x=\frac{8(sin(51))}{sin(90}$

Which can simplify to:

$x=8(sin(51))$

Which putting into a calculator gets you:

$x=6.2$

So the line PR is 6.2 units long.

6 0

3 years ago

If 4 Maths books are selected from 6 different Maths different English books, how many ways can the seven om Maths books and 3 E

alexgriva [62]

Answer:

Please see the answer below

Step-by-step explanation:

a. Since there’s no restrictions .Therefore , the number of ways = 7!*150 = 756000

b. The number of ways such that the 4 math books remain together

The pattern is as follows: MMMMEEE, EMMMMEE, EEMMMME, and EEEMMMM

Where M = Math’s Book and E= English Book.

Number of ways = 4!*8!*4*150= 86400 ways.

c. The number of ways such that math book is at the beginning of the shelf

The number of ways = 6!*4*150 = 432000

d. The number of ways such that math and English books alternate

The number of ways = 150*4!*3! =2160 ways

e. The number of ways such that math is at the beginning and an English book is in the middle of the shelf. The number of ways = 4*3*5!*150 =216000 ways.

6 0

3 years ago

find two consecutive odd integers that 7 times the smaller integer is 61 more than 2 times the larger integer.

Aleks04 [339]

Answer:

x = 13 smaller integer

x + 2 = 15 the other integer

Step-by-step explanation:

x = the smaller integer

x + 2 = the next integer

7(x) = 61 + 2(x + 2)

7x = 61 + 2x +4

5x = 61 + 4

5x = 65

x = 13 smaller integer

x + 2 = 15 the other integer

8 0

2 years ago