Answer:

Step-by-step explanation:
![\displaystyle a^2 + 2a\sqrt{\frac{a^2}{4} + h^2} = S.A. \\ \\ 14^2 + 2[14]\sqrt{\frac{14^2}{4} + 19^2} = S.A. \\ \\ 196 + 28\sqrt{\frac{196}{4} + 361} ≈ 762,9567885 ≈ 762,96](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%5E2%20%2B%202a%5Csqrt%7B%5Cfrac%7Ba%5E2%7D%7B4%7D%20%2B%20h%5E2%7D%20%3D%20S.A.%20%5C%5C%20%5C%5C%20%2014%5E2%20%2B%202%5B14%5D%5Csqrt%7B%5Cfrac%7B14%5E2%7D%7B4%7D%20%2B%2019%5E2%7D%20%3D%20S.A.%20%5C%5C%20%5C%5C%20196%20%2B%2028%5Csqrt%7B%5Cfrac%7B196%7D%7B4%7D%20%2B%20361%7D%20%E2%89%88%20762%2C9567885%20%E2%89%88%20762%2C96)
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Answer:
(5 , 3)
Step-by-step explanation:
x + y = 8 ----------------(I)
-x + 3y = 4 ----------(II)
Add equation (I) and (II) and so x will be eliminated and we can find the value of y
(I) x + y = 8
(II) -<u>x + 3y = 4 </u>
4y = 12 {Divide both sides by 4}
y = 12/4
y = 3
Plugin y = 3 inn equation (I)
x + 3 = 8 {Subtract 3 from both sides}
x = 8 -3
x = 5
Answer:
a) 
b) 
Step-by-step explanation:
We know that theres a <u><em>linear relationship</em></u> between the rate of the chirping of crickets and the air temperature.
<u><em>The equation of a line is</em></u>:

So, let's name our variables
x= Temperature
y=Rate of the chirping
First of all, we need to find the <u><em>slope</em></u> with the two given points
60ºF , 
80ºF, 
By,


<u>Now, the </u><em><u>equation</u></em><u> between the air temperature and the number of chirps is:</u>


Solving for y,
a) 
b) <u>To calculate the </u><u>rate</u><u> at which the crickets chirp when the </u><u>temperature is 102 ºF</u><u> we need to evaluate y(102)</u>
