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Viefleur [7K]
3 years ago
12

PLEASE HELP..........................................

Mathematics
1 answer:
Scrat [10]3 years ago
3 0
24 units
9+9+3+3 =24.
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What is 9999 times 886 + 234
barxatty [35]

9999*886+234

(9999*886)+234

= 8859114+234

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I hope that's help !

7 0
3 years ago
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Seven more than one half a number as an equation
romanna [79]

Answer:

1/2x+7

Step-by-step explanation:

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3 years ago
It took Fred 17 days to save $36.55. At that rate, how long would it take him to save $70.95?
Vlad1618 [11]

Answer:

Fred would take 33 days to save $ 70.95.

Step-by-step explanation:

The number of days (t) is directly proportional to the quantity of money saved (c), in monetary units, then we can calculate the time taken to save $ 70.95 by simple rule of three:

t = 17\,d\,\times \frac{\$\,70.95}{\$\,36.55}

t = 33\,d

Fred would take 33 days to save $ 70.95.

4 0
2 years ago
Given that g(x)=x-3/x+4 find each of the following.
maks197457 [2]

Given that the function g(x)=x-3/x+4, the evaluation gives:

  1. g(9) = 6/13.
  2. g(3) = 0.
  3. g(-4) = undefined.
  4. g(-18.75) = 1.07.
  5. g(x+h) = x+h-3/x+h+4

<h3>How to evaluate the function?</h3>

In this exercise, you're required to determine the value of the function g at different intervals. Thus, we would substitute the given value into the function and then evaluate as follows:

When g = 9, we have:

g(x)=x-3/x+4

g(9) = 9-3/9+4

g(9) = 6/13.

When g = 3, we have:

g(x)=x-3/x+4

g(3) = 3-3/3+4

g(3) = 0/13.

g(3) = 0.

When g = -4, we have:

g(x)=x-3/x+4

g(-4) = -4-3/-4+4

g(-4) = -1/0.

g(-4) = undefined.

When g = -18.75, we have:

g(x)=x-3/x+4

g(-18.75) = -18.75-3/-18.75+4

g(-18.75) = -15.75/-14.75.

g(-18.75) = 1.07.

When g = x+h, we have:

g(x)=x-3/x+4

g(x+h) = x+h-3/x+h+4

Read more on function here: brainly.com/question/17610972

#SPJ1

8 0
2 years ago
What is the value of j(-5)
elena55 [62]

f(x)=\left\{\begin{array}{ccc}2-x&if&x\leq-5\\x+11&if&x > -5\end{array}\right\\\\-5\leq-5\ \text{therefore for}\ x=-5,\ f(x)=2-x\\\\f(-5)=2-(-5)=2+5=7

6 0
3 years ago
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