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2 years ago

The area of a square is 50 square yards. What is the length, in yards, of one side of the square?

Mathematics

1 answer:

Elena L [17]2 years ago

8 0

Answer:

$area \: of \: a \: square = {x}^{2} \\ here \: {x}^{2} = 50 \\ x = \sqrt{50} = \sqrt{2 \times 25} = 5 \sqrt{2} \\ one \: side = 5 \sqrt{2} yard$

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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

Which expression is equivalent to the product of p+7/3 and 6/p , where p is not equal to 0?

vichka [17]

Answer:

$\frac{(6p+14)}{p}$

Step-by-step explanation:

$(p+\frac{7}{3})(\frac{6}{p})$

Making denominator same in first bracket we get

$(\frac{3p+7}{3})(\frac{6}{p})$

$(\frac{(3p+7)*6}{3*p})$

Dividing 6 by 3 we get 2

$(\frac{(3p+7)*2}{p})$

using distributive law

$(\frac{6p+14}{p})\\$

Hence this is our answer

8 0

2 years ago

Read 2 more answers

Which two expressions each represent 3/12

tester [92]

Answer:

1/4 or 25% Your Welcome!

7 0

3 years ago

Demi and Juan start saving money for their vacation. Demi starts with $50 and saves $20 each week; Juan started with only $35, b

Fiesta28 [93]

Answer:

i d k

Step-by-step explanation:

7 0

2 years ago

równanie zmiany prędkość autokaru poruszającego się po prostym odcinku szosy i rozpoczynającego hamowanie od szybkości 20m/s ma

Rasek [7]

Answer:

s = 22.5 m

Step-by-step explanation:

the equation for the speed change of a coach moving along a straight section of the road and starting braking at a speed of 20 m / s has the form v (t) = 25-5t. Using integral calculus, determine the coach's braking distance.

v (t) = 25 - 5 t

at t = 0 , v = 20 m/s

Let the distance is s.

$s =\int v(t) dt\\\\s =\int (25 - 5t)dt\\\\s= 25 t - 2.5 t^2 \\$

Let at t = t, the v = 20

So,

20 = 25 - 5 t

t = 1 s

So, s = 25 x 1 - 2.5 x 1 = 22.5 m

3 0

2 years ago