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TiliK225 [7]
3 years ago
9

Guys please help me !! ASAP !

Mathematics
2 answers:
omeli [17]3 years ago
5 0

Answer: 1. vertical

2. alternate interior

3. corresponding

4.corresponding

5. adjacent

Step-by-step explanation:

jenyasd209 [6]3 years ago
4 0
1. vertical
2. alternate interior
3. corresponding
4.corresponding
5. adjacent
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Of your 32 classmates, 1/3
ozzi

Answer:

C.1/2

Step-by-step explanation:

1/3 have siblings and 1/2 of those are girls

1/2 of the siblings are girls

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3 years ago
What type(s) of symmetry does the figure have?
Troyanec [42]

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both points and rotational only

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3 years ago
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On sunday 1/16 people who visit were senior citizens, 3/16 were infants, 3/8 were children and 3/8 were adults, how many of each
mamaluj [8]

Answer:

Out of total people visited the zoo on Sunday , 6.25% was senior citizen ,  18.75% were infants , 35% were children and 35% were adult.

Step-by-step explanation:

let total number of people visited the group  = x

so number of senior citizens = x × 1/16 = x/16

Percentage of  senior citizens = (number of senior citizens ÷ total number of people visited the group) × 100

= (x/16 ÷ x)  × 100 = 100/16 = 6.25%

number of infants = 3/16 × x = 3x/16

Percentage of  infants =(number of infants  ÷ total number of people visited the group) × 100 = (3x/16 ÷ x)  × 100 = 300/16 = 18.75%

number of  children = 3/8 × x = 3x/8

Percentage of  children =(number of children ÷ total number of people visited the group) × 100 = (3x/8 ÷ x)  × 100 = 300/8 = 37.5%

number of  adults = 3/8 × x = 3x/8

Percentage of  adults =(number of adults ÷ total number of people visited the group) × 100 = (3x/8 ÷ x)  × 100 = 300/8 = 37.5%.

Hence we can conclude that Out of total people visited the zoo on Sunday , 6.25% was senior citizen ,  18.75% were infants , 35% were children and 35% were adult.






4 0
3 years ago
A certain car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year
icang [17]

Answer:

b(b/a)^2

Step-by-step explanation:

Given that the value of the car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year and that car was worth a dollars on December 31, 2010 and was worth b dollars on December 31, 2011, then

b = a - (p% × a) = a(1-p%)

b/a = 1 - p%

p% = 1 - b/a = (a-b)/a

Let the worth of the car on December 31, 2012 be c

then

c = b - (b × p%) = b(1-p%)

Let the worth of the car on December 31, 2013 be d

then

d = c - (c × p%)

d = c(1-p%)

d = b(1-p%)(1-p%)

d = b(1-p%)^2

d = b(1- (a-b)/a)^2

d = b((a-a+b)/a)^2

d = b(b/a)^2 = b^3/a^2

The car's worth on December 31, 2013 =  b(b/a)^2 = b^3/a^2

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What is 679 rounded to the nearest ten
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680 is the answer to that question
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