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3 years ago

Absolute value to show the distance between -70 and 15

Mathematics

1 answer:

castortr0y [4]3 years ago

7 0

To find the absolute value, change all into positives, and add them

|-70| + 15 = answer

|-70| = 70

70 + 15 = 85

The distance in the number line between -70 and 15 is 85

85 is your answer

hope this helps

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vodka [1.7K]

Answer:

2 1/4

Before Simplifying:

9/4

After Simplifying:

2 1/4

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3 years ago

Read 2 more answers

mike's condo has a market value of 310000. the property in mikes area is assessed at 40% of the market value. the tax rate is 14

Svetradugi [14.3K]

If you would like to know Mike's tax, you can calculate this using the following steps:

a market value ... 310000
the property ... 40% of the market value = 40% * 310000 = 40/100 * 310000 = 124000

the tax rate ... 145.10 per 1000 of assessed valuation
145.10 ... 1000 of assessed valuation
x ... 124000 of assessed valuation
____________________________
145.10 * 124000 = 1000 * x /1000
x = 145.10 * 124000 / 1000
x = 17992.4 (Mike's tax)

The correct result would be: Mike's tax is 17992.4.

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3 years ago

A) Find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s.

Fed [463]

Answer:

A) $a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}$

B) $a_{0} = a_{1} = 0$

C) for n = 2

$a_{2}$ = 1

for n = 3

$a_{3}$ = 3

for n = 4

$a_{4}$ = 8

for n = 5

$a_{5}$ = 19

Step-by-step explanation:

A) A recurrence relation for the number of bit strings of length n that contain a pair of consecutive Os can be represented below

if a string (n ) ends with 00 for n-2 positions there are a pair of consecutive Os therefore there will be : $2^{n-2}$ strings

therefore for n ≥ 2

The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os

$a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}$

b ) The initial conditions

The initial conditions are : $a_{0} = a_{1} = 0$

C) The number of bit strings of length seven containing two consecutive 0s

here we apply the re occurrence relation and the initial conditions

$a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}$

for n = 2

$a_{2}$ = 1

for n = 3

$a_{3}$ = 3

for n = 4

$a_{4}$ = 8

for n = 5

$a_{5}$ = 19

7 0

3 years ago

A wedding at Boone Hall Plantation costs $22,200 for 120 people. How much does Boone Hall charge per guest?

alina1380 [7]

Answer:

$185 per person

Step-by-step explanation:

22,200/120=185

8 0

3 years ago

How do I begin this equation

alex41 [277]

Trigonometry:
We know that csc x = 1 / sin x and π = 180°;
sin 5π/ 6 = sin 150° = 1/2
csc 2π / 3 = 1 / sin 120° = 1 / √3/2 = 2 / √3
Finally:
4 · 1/2 + 7 · ( 2 / √3 )² = 2 + 7 · 4/3 =
= 2 + 28/3 = 2 + 9 1/3 = 11 1/3

5 0

4 years ago