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Karolina [17]
3 years ago
15

What is the value of log^416

Mathematics
2 answers:
crimeas [40]3 years ago
6 0

Answer:

log 4

Step-by-step explanation:

Log 4 is the answer

alukav5142 [94]3 years ago
6 0

Answer: 4

Step-by-step explanation: hope this helps

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2x^2 + 4x +7 quadratic formula
marissa [1.9K]

Answer:

Step-by-step explanation:

7 0
2 years ago
Please show all your work. Ms. Lee is now planning a square flower garden as a memorial. She is expanding an existing flower gar
Lynna [10]
169 needs to be into four parts, by which 2 sides multiply to 169. The sides would be 13 each, that way 13×13=169. You would then subtract 6m from the 13m, which gives you 7 as your answer!

Answer: 7m
3 0
3 years ago
Daniel wants to have a 90 average in his math class at the end of the year. He is trying to determine what he needs to get on hi
morpeh [17]
Exam + Tests + Quizzes + HW + Projects = 90

Exam = 10% * grade (x) = 0.1 * x 

Tests = 50% * 85 = 0.5 * 85 = 42.5

Quizzes  = 15% * 95 = 0.15 * 95 = 14.25

HW = 15% * 98 = 0.15 * 98 = 14.7

Projects = 10% * 92 = 0.1 * 92 = 9.2

Exam + Tests + Quizzes + HW + Projects = 90

(0.1 * x) + 42.5 + 14.25 + 14.7 + 9.2 = 90

(0.1 * x) + 80.65 = 90      
  (subtract 80.65 from each side)

(0.1 * x) = 9.35        
  (divide 0.1 from each side)

(0.1 * x)/0.1 = 9.35/0.1

x = 0.935

The answer is 93.5%


4 0
3 years ago
Read 2 more answers
Find F"(x) if f(x) = cot (x)
hammer [34]

f(x)=\cot x\implies f'(x)=-\csc^2x\implies\boxed{f''(x)=2\csc^2x\cot x}

If you don't know the first derivative of \cot, but you do for \sin and \cos, you can derive the former via the quotient rule:

\cot x=\dfrac{\cos x}{\sin x}

\implies(\cot x)'=\dfrac{\sin x(-\sin x)-\cos x(\cos x)}{\sin^2x}=-\dfrac1{\sin^2x}=-\csc^2x

or if you know the derivative of \tan:

\cot x=\dfrac1{\tan x}

\implies(\cot x)'=-(\tan x)^{-2}\sec^2x=-\dfrac{\sec^2x}{\tan^2x}=-\dfrac{\frac1{\cos^2x}}{\frac{\sin^2x}{\cos^2x}}=-\dfrac1{\sin^2x}=-\csc^2x

As for the second derivative, you can use the power/chain rules:

(-\csc^2x)'=-2\csc x(\csc x)'=-2\csc x(-\csc x\cot x)=2\csc^2x\cot x

or if you don't know the derivative of \csc,

\csc x=\dfrac1{\sin x}

\implies(-\csc^2x)'=\left(-(\sin x)^{-2}\right)'=2(\sin x)^{-3}(\sin x)'=\dfrac{2\cos x}{\sin^3x}

which is the same as the previous result since

\csc^2x\cot x=\dfrac1{\sin^2x}\dfrac{\cos x}{\sin x}=\dfrac{\cos x}{\sin^3x}

4 0
3 years ago
A plant inspector takes a random sample of six month old seedlings from a nursery and measures their heights to the nearest mm .
morpeh [17]

Answer:

48

Step-by-step explanation:

3 0
2 years ago
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