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vodomira [7]
2 years ago
13

SOMEONE, PLEASE HELP LITERALLY STRUGGLING!!!!!!

Mathematics
2 answers:
givi [52]2 years ago
5 0
I think it is A . hope it helps
Sonbull [250]2 years ago
3 0

Answer:

A

Step-by-step explanation:

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Danielle has $3 in quarters and $2 in dimes in her pocket. If she randomly selects one coin out of her
sattari [20]
The answer is C. 3/5
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Factor 5a^2-30a+45, and 2x^5+x^4-2x-1.
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Step-by-step explanation:

{5a}^{2}  - 30a + 45

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5(a-3)^2

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Lubov Fominskaja [6]

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7 0
3 years ago
Read 2 more answers
Divide 3.7÷0.006 round the answer the nearest hundredth?
Julli [10]
First divide 3.7 by 0.006
Which is 616.666
The last 6 is the thousandths place. If it is 5 or bigger then the number to the left is rounded up. If its lower than 5 you round down

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6 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

4 0
3 years ago
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