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vaieri [72.5K]
3 years ago
13

Can anyone help me with this

Mathematics
1 answer:
mrs_skeptik [129]3 years ago
8 0
Suzanne will C use 610 stickers and spend D. 13.70
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Answer:

a) I(95,50) = 73.19 degrees

b) I_{T}(95,50) = -7.73

Step-by-step explanation:

An approximate formula for the heat index that is valid for (T ,H) near (90, 40) is:

I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}

a) Calculate I at (T ,H) = (95, 50).

I(95,50) = 45.33 + 0.6845*(95) + 5.758*(50) - 0.00365*(95)^{2} - 0.1565*95*50 + 0.001*50*95^{2} = 73.19 degrees

(b) Which partial derivative tells us the increase in I per degree increase in T when (T ,H) = (95, 50)? Calculate this partial derivative.

This is the partial derivative of I in function of T, that is I_{T}(T,H). So

I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}

I_{T}(T,H) = 0.6845 - 2*0.00365T - 0.1565H + 2*0.001H

I_{T}(95,50) = 0.6845 - 2*0.00365*(95) - 0.1565*(50) + 2*0.001(50) = -7.73

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If a bike is on sale for 45% off of the regular price. If the sale price is $299.75 what was the original price
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