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katrin [286]
4 years ago
7

Can SOMEONE HELP ASAP

Mathematics
1 answer:
Hunter-Best [27]4 years ago
3 0
Y > -x - 3
y = -x - 3
for x = 0 → y = -0 - 3 = -3 → (0, -3)for x = -3 → y = -(-3) - 3 = 3 - 3 = 0 → (-3, 0)
y ≤ 4x + 2
y = 4x + 2
for x = 0 → y = 4 · 0 + 2 = 2 → (0, 2)for x = -1 → y = 4 · (-1) + 2 = -4 + 2 = -2 → (-1, -2)
Answer is in attachment.

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What strategy can be used to solve 8 x 25 x 23
wariber [46]
The best way to multiply complicated problems mentally* is to break them into chunks.

Start with 20 times 8 to get 160, and then 8 times 5 to get 40. Add these together to get 200. 200 times 3 is 600, and 20 times 200 is 4000. Add these together to get your final answer, 4600. 

In other words 8 x 20 + 8 x 5 = 200  200 x 3 + 200 x 20 = 4600

*I assumed you meant mentally, because otherwise you'd just multiply longhand, multiplying two of the three numbers and then multiplying their product with the third. 
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Read 2 more answers
The heat index I is a measure of how hot it feels when the relative humidity is H (as a percentage) and the actual air temperatu
PSYCHO15rus [73]

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

8 0
3 years ago
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