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Maksim231197 [3]
3 years ago
9

Four equivalent expressions of 36 mince 9

Mathematics
2 answers:
Vlad [161]3 years ago
4 0
36-9
-9+36
1 (36-9)
36-(3×3)
soldi70 [24.7K]3 years ago
4 0
The answer is 27. Making 30 - 3, 9 x 3, 20 + 7, and 25 + 2 all valid. 
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Look at triangle ABC. What is the length of said AB of the triangle
Nuetrik [128]
\bf ~~~~~~~~~~~~\textit{distance between 2 points}
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A(\stackrel{x_1}{4}~,~\stackrel{y_1}{5})\qquad 
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d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
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AB=\sqrt{(1-4)^2+(2-5)^2}\implies AB=\sqrt{(-3)^2+(-3)^2}
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AB=\sqrt{9+9}\implies \boxed{AB=\sqrt{18}}\implies  AB=\sqrt{2(9)}
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5 0
3 years ago
Your child is prescribed amoxicillin (40mg/kg/day; not to exceed 1500 mg per 24 hour period) to be
Reptile [31]

Answer:

491 mg per dose

Step-by-step explanation:

1 kg ≈ 2.2 lbs

81/2.2 36.82 kg

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to the nearest mg = 491 mg per dose

5 0
2 years ago
I need help again please
babymother [125]
There are 8 pints in a gallon. It will take 10 days to drink 1 5-gallon container.
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3 0
3 years ago
Read 2 more answers
What is the degree of the monomial -2<br> show all work!
Leni [432]

-2 is a degree 0 polynomial with 1 term and 0 variables

3 0
3 years ago
Use Newton's Method to approximate the zero(s) of the function. Continue the iterations until two successive approximations diff
liubo4ka [24]

Answer:

There is only one real zero and it is located at x = 1.359

Step-by-step explanation:

After the 4th iteration the solution was repeating the first 3 decimal places.  The formula for Newton's Method is

x_{n}-\frac{f(x_{n}) }{f'(x_{n}) }

If our function is

f(x)=x^5+x-6

then the first derivative is

f'(x)=5x^4+1

I graphed this on my calculator to see where the zero(s) looked like they might be, and saw there was only one real one, somewhere between 1 and 2.  I started with my first guess being x = 1.

When I plugged in a 1 for x, I got a zero of 5/3.  

Plugging in 5/3 and completing the process again gave me 997/687

Plugging in 997/687 and completing the process again gave me 1.36976

Plugging in 1.36976 and completing the process again gave me 1.359454

Plugging in 1.359454 and completing the process again gave me 1.359304

Since we are looking for accuracy to 3 decimal places, there was no need to go further.

Checking the zeros on the calculator graphing program gave me a zero of 1.3593041 which is exactly the same as my 5th iteration!

Newton's Method is absolutely amazing!!!

5 0
3 years ago
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