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Arada [10]
3 years ago
9

the human eyeball is a sphere with a surface area of 20 square cm. find the approximate radius of the human eyeball.

Mathematics
1 answer:
guapka [62]3 years ago
4 0

Answer:

1.26 corect answer

Step-by-step explanation:

the surface area of a sphere is 4*pi*r^2  

so r is sqrt(20/pi) which is sqrt(1.59) = 1.26

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PLS HURRY, I neeeedd helppppp
Sedaia [141]

Answer:

third option

r > 11

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hope it helps

4 0
2 years ago
Write the prime factorization of 12. Use exponents when appropriate and order the factors from least to greatest (for example,22
KatRina [158]

Answer:

the answer is 2×2×3 I think

6 0
3 years ago
the price of the cell phone is expected to increase by 165% in September 2015. The current price of a cell phone is $62.80. What
Sladkaya [172]

Answer:

$166.42.

Step-by-step explanation:

We have been given that the price of the cell phone is expected to increase by 165% in September 2015. The current price of a cell phone is $62.80.

The price of cell phone in September 2015 will be $62.80 plus 165% of $62.80.

\text{The predicted cost of a cell phone in September 2015}=\$62.80+(\frac{165}{100}*\$62.80)

\text{The predicted cost of a cell phone in September 2015}=\$62.80+(1.65*\$62.80)

\text{The predicted cost of a cell phone in September 2015}=\$62.80+\$103.62

\text{The predicted cost of a cell phone in September 2015}=\$166.42

Therefore, the predicted cost of a cell phone in September 2015 will be $166.42.

4 0
2 years ago
Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to f
tino4ka555 [31]

Answer:

The minimum value of the given function is f(0) = 0

Step-by-step explanation:

Explanation:-

Extreme value :-  f(a, b) is said to be an extreme value of given function 'f' , if it is a maximum or minimum value.

i) the necessary and sufficient condition for f(x)  to have a maximum or minimum at given point.

ii)  find first derivative f^{l} (x) and equating zero

iii) solve and find 'x' values

iv) Find second derivative f^{ll}(x) >0 then find the minimum value at x=a

v) Find second derivative f^{ll}(x) then find the maximum value at x=a

Problem:-

Given function is f(x) = log ( x^2 +1)

<u>step1:</u>- find first derivative f^{l} (x) and equating zero

  f^{l}(x) = \frac{1}{x^2+1} \frac{d}{dx}(x^2+1)

f^{l}(x) = \frac{1}{x^2+1} (2x)  ……………(1)

f^{l}(x) = \frac{1}{x^2+1} (2x)=0

the point is x=0

<u>step2:-</u>

Again differentiating with respective to 'x', we get

f^{ll}(x)=\frac{x^2+1(2)-2x(2x)}{(x^2+1)^2}

on simplification , we get

f^{ll}(x) = \frac{-2x^2+2}{(x^2+1)^2}

put x= 0 we get f^{ll}(0) = \frac{2}{(1)^2}   > 0

f^{ll}(x) >0 then find the minimum value at x=0

<u>Final answer</u>:-

The minimum value of the given function is f(0) = 0

5 0
3 years ago
2.35 ÷ (-1.5) (Round to the tenth)
Sidana [21]
-1.6 is your answer :))))
5 0
3 years ago
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