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SSSSS [86.1K]
3 years ago
12

please Order least to greatest -2 3/10, -2 2/5, -2 -2 1/2, -3 And yes those are negative signs

Mathematics
2 answers:
rewona [7]3 years ago
5 0
-3; -2 3/10; -2 2/5; -2 1/2

Your welcome!
galina1969 [7]3 years ago
5 0
-3, -2 1/2 , -2 2/5, - 2 3/10, -2
This is the answer because you should find out the common denominator (Hint hint nudge nudge wink wink it is 10..) Then you order it from least to greatest.
~JZ
Hope it helps
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What is lim x → 9+ ( x - 9/| 9 - x | )
SIZIF [17.4K]

Answer:

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Step-by-step explanation:

6 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Mama L [17]

(a-b)^2=a^2-2ab+b^2

x^2-10x+27=0\ \ \ \ |-27\\\\x^2-10x=-27\\\\x^2-2\cdot x\cdot5+27=0\ \ \ |+5^2=25\\\\x^2-10x+25=-27+25

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4(x+3)-(2x-1)=30+7x-1+5x
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3 years ago
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andre [41]
Answer: Choice A) $4500.33

---------------------------------------------------------------------

Explanation:

The formula you'll use is
F = P*(1+r)^t

where
F = final amount
P = initial amount
r = growth rate (in decimal form)
t = time in years

In this case,
F = unknown (this is what we're trying to figure out)
P = 3046 
r = 0.05 (since 5% = 5/100 = 0.05)
t = 8

Plug those three known values into the formula and evaluate

F = P*(1+r)^t
F = 3046*(1+0.05)^8
F = 3046*(1.05)^8
F = 3046*1.4774554
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4 0
2 years ago
Help ASAP, i don't understand this.
Inessa [10]

The  given polynomial expressed as partial fraction is  5x^3 + 9/x

<h3>Polynomial function and partial fractions</h3>

Polynomial function are function that have a leading degree of 3. Given the expression below;

f(x) = (5x^4+9)/x

Applying partial fraction

Since the part fraction means separating the fraction into two parts, hence;

f(x) = 5x^4/x + 9/x

Simplify

Since there is value of x at both numerator and denominator, it will cancel out to have

f(x) = 5x^3 + 9/x

From the result, we can see that p(x) = 5x^3 and k = 9

Hence the  given polynomial expressed as partial fraction is  5x^3 + 9/x

Learn more on polynomial here: brainly.com/question/2833285

#SPJ1

8 0
2 years ago
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