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fredd [130]
3 years ago
13

Rani had Rs800 left after spending 75% of her money.How much money did she have in the beginning?

Mathematics
1 answer:
hram777 [196]3 years ago
6 0

Answer:

  • Rs. 3200

Step-by-step explanation:

Money left Rs. 800

Money spent = 75%

  • Rs 800 = 25%
  • 800 = 0.25x
  • x = 800/0.25
  • x = 3200

Initial amount was Rs. 3200

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Please help. I don't understand how to solve this problem.
spayn [35]

Answer:

BF=16

Step-by-step explanation:

To find BF, (I will be calling it x) you need to use the equation

CF/FB=CE/EA Substitute

FB=x

24/x=18/12 cross multiply

18x=288 divide both sides by 18

x=16

FB=16

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3 0
3 years ago
What is 2x2+2 mmmmmmmmmmmmmmmmmmm
Amanda [17]

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6

Step-by-step explanation:

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7 0
2 years ago
Let f (x) = 2x - 1, g(x) = 3x, and h(x) = x2 + 1. Compute the following: h (h(x)) and. g (h (x))
ki77a [65]
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h(h(x)) = (x²+1)² + 1
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= (x^4 + x² + x² + 1) + 1 
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7 0
2 years ago
What is lim x→3 x^2 - x - 6/ x - 3
Kisachek [45]
<h3>Answer: D) 5</h3>

=======================================================

Explanation:

If we plugged x = 3 into the expression, then we'd get x-3 = 3-3 = 0 in the denominator. That's not allowed. But we can simplify first

x^2-x-6 factors to (x-3)(x+2). The key here is that (x-3) is a factor. It cancels with the x-3 in the denominator

So, \frac{x^2-x-6}{x-3} = \frac{(x-3)(x+2)}{x-3} = x+2

Allowing us to say,

\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = \lim_{x \to 3}\frac{(x-3)(x+2)}{x-3}\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = \lim_{x \to 3}(x+2)\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = 3+2\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = 5\\\\\\

5 0
3 years ago
What is 10 radical 12 times 6 radical 6
romanna [79]

Answer:

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Step-by-step explanation:

simplify the radicals

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calculate the products

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= 120 radical 18

again simplify

360 radical 2  

8 0
2 years ago
Read 2 more answers
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