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horrorfan [7]
3 years ago
12

How old would I be if I was born in 1921

Mathematics
1 answer:
earnstyle [38]3 years ago
6 0
2016 - 1921 = 95
Hope this helped!
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The function f is continuous on the interval [3, 13] with selected values of x and f(x) given in the table below. Find the avera
Lelechka [254]

The average rate of change of f(x) over the interval [3, 13] is given by the ratio

\dfrac{f(13)-f(3)}{13-3}=\dfrac{22-2}{13-3}=\dfrac{20}{10}=2

3 0
3 years ago
2.9 x 10^5 + 8.7 x 10^5
LiRa [457]

Answer:

So this is scientific notation what you do is

2.9x10^5 so you put however many zeros the is powered to 10. See there is a 5 It has to be behind or in front of the decimal you might be saying how do you know if its behind or in front well negative is in front and positive is behind .

<u><em>290,000 </em></u>there was a number already behind the decimal so there you go 5 number behind the decimal.  

Same for the second one

<u><em>290,000</em></u><em>  </em><u><em>870,000</em></u> is your answers so now you add hem together

<em><u>1,160,000 is your final answer </u></em>

Step-by-step explanation:

3 0
3 years ago
John is arguing that they should use the following expressions to calculate the final cost of this item: (20-20•0.15)+(20-20•0.1
Vikki [24]
If they use (20-20x0.15)+(20-20x.015)x0.05 the final cost would be $34.05

Hope this helps

3 0
4 years ago
*EASY* Solve with steps! *WILL MARK BRAINLIEST*
SpyIntel [72]

Answer:

OK I will If u need any math help just ask me!!

Step-by-step explanation:

Have a wonderful day!!!

7 0
3 years ago
Read 2 more answers
A restaurant manager can spend at most $600 a day for operating costs and payroll. It costs $100 each day to operate the bank an
navik [9.2K]

Option A

The restaurant Manager can afford at most 10 employees for the day

<em><u>Solution:</u></em>

Given that restaurant manager can spend at most $600 a day for operating costs and payroll

It costs $100 each day to operate the bank and $50 dollars a day for each employee

The given inequality is:

50x + 100\leq 600

Where , "x" is the number of employees per day

Let us solve the inequality for "x"

50x + 100\leq 600

Add -100 on both sides of inequality

50x + 100 - 100\leq 600 - 100\\\\50x\leq 500

Divide by 50 on both sides of inequality

\frac{50x}{50}\leq \frac{500}{50}\\\\x\leq 10

Hence the restaurant Manager can afford at most 10 employees for the day

Thus option A is correct

6 0
4 years ago
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