3 years ago

How do I graph a solution set for the inequality 300+5x>550?

1 answer:

3 0

First to solve the inequality
300 + 5x > 550
5x > 550 - 300
5x > 250
x > 50

Therefor the answers are all numbers bigger than 50

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patriot [66]

lets solve for x
flip the equation
add-4 to both sides
divide both sides by 3
x=1/3y+-4/3

4 0

3 years ago

liraira [26]

1.=9900
2.=9.3
3.= x= 12.2155
Hope this helps

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3 years ago

faust18 [17]

Answer:

im not the best but i think 200 or 400 m

Step-by-step explanation:

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3 years ago

Lelu [443]

$7.5\% \text { of } x = 9 \text { students}$

$0.75 \times x = 9 \text { students}$

$0.75x = 9$

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3 years ago

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