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pickupchik [31]
3 years ago
15

A medical office is replacing their existing flooring with carpeting in the waiting and office areas and laminate flooring in th

e patient exam rooms. If each of the 10 exam rooms is 10 feet by 10 feet, the waiting area is 20 feet by 15 feet, and the office area is 25 by 20 feet, how many total square feet of flooring will they be replacing?
Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
8 0
Area of 1 exam room = 10 ft. × 10 ft. = 100 ft.²
Area of 10 exam room = 100 ft.² × 10 = 1000 ft.²

Area of waiting room = 20 ft. × 15 ft. = 300 ft.²

Area of office area = 25 ft. × 20 ft. = 500 ft.²

Now,
Total area of the hospital to be replaced = 1000 ft.² + 300 ft.² + 500 ft.² = 1800 ft.²
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Another teacher has a monthly salary of $2000 She spends $943 per month on rent Her monthly salary then increases by 15% What pe
Aleks04 [339]

Answer:

47.15%

Step-by-step explanation:

Getting a raise of 15% would increase the teacher's salary by 300. 943 of 2300 is 47.15%.

5 0
2 years ago
For the reaction 2Fe + O2 → 2FeO, how many grams of iron oxide are produced from 42.4 mol of iron?
Butoxors [25]

In the balanced reaction

2 Fe + O₂   →   2 FeO

2 moles of iron (Fe) react with 1 mole of molecular oxygen (O₂) to produce 2 moles of iron oxide (FeO). The ratio of Fe to FeO is 1-to-1, so if one starts with 42.4 mol of Fe, one will end up with the same amount of FeO, 42.4 mol.

Look up the molar mass of Fe and O:

• Fe = 55.845 g/mol

• O = 15.999 g/mol

Then the molar mass of FeO is approximately 71.835 g/mol, and so the mass of 42.4 mol of FeO is

(42.4 mol) × (71.835 g/mol) ≈ 3050 g

8 0
2 years ago
How do you do 137 ÷ 7 in a area model
aliya0001 [1]
Hey,

If you were to divide 137/7 it would be it would be 19 or if you wont a decimal it would be 19.5

I Hoped that Helped 
8 0
2 years ago
Read 2 more answers
The timekeeper for a particular mile race uses a stopwatch to determine the finishing times of the racers. He then calculates th
insens350 [35]

Answer:

300 seconds = 5 minutes

Step-by-step explanation:

At first convert the time in minutes which is

5.75 * 60 = 345 seconds

No as the average is 345 seconds at starts from 45 so we subtract that time from 45 seconds

345-45 = 300 seconds (Mean)

So 300 seconds = 300/60 = 5 minutes

So the Accurate mean is 5 minutes

7 0
3 years ago
Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|
Fynjy0 [20]

Answer:

R3 <= 0.083

Step-by-step explanation:

f(x)=xlnx,

The derivatives are as follows:

f'(x)=1+lnx,

f"(x)=1/x,

f"'(x)=-1/x²

f^(4)(x)=2/x³

Simialrly;

f(1) = 0,

f'(1) = 1,

f"(1) = 1,

f"'(1) = -1,

f^(4)(1) = 2

As such;

T1 = f(1) + f'(1)(x-1)

T1 = 0+1(x-1)

T1 = x - 1

T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2

T2 = 0+1(x-1)+1(x-1)^2

T2 = x-1+(x²-2x+1)/2

T2 = x²/2 - 1/2

T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3

T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3

T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)

Thus, T1(2) = 2 - 1

T1(2) = 1

T2 (2) = 2²/2 - 1/2

T2 (2) = 3/2

T2 (2) = 1.5

T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)

T3(2) = 4/3

T3(2) = 1.333

Since;

f(2) = 2 × ln(2)

f(2) = 2×0.693147 =

f(2) = 1.386294

Since;

f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).

Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,

Since;

f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2

Finally;

R3 <= |2/(4!)(2-1)^4|

R3 <= | 2 / 24× 1 |

R3 <= 1/12

R3 <= 0.083

5 0
2 years ago
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