Mathematics, 02.01.2021 17:00 lizreed2002

During a certain24 -hour period, the temperature at time t (measured in hours from the start of the period) was T(t)=50+6t-1/3t^2 degrees. What was the average temperature during that period?

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Answer from: caleb768

The value of average temperature during that period= $58^{\circ}$

Step-by-step explanation:

We are given that

Temperature at time t is given by

$T(t)=50+6t-\frac{1}{3}t^2$

We have to find the average temperature during a certain 24-hour period.

We know that

Average value of function

$f_{av}=\frac{1}{b-a}\int_{a}^{b}f(x)dx$

Using the formula

Average temperature during the period

$=\frac{1}{24-0}\int_{0}^{24}(50+6t-\frac{1}{3}t^2)dt$

$=\frac{1}{24}[50t+3t^2-\frac{1}{9}t^3]^{24}_{0}$

$=\frac{1}{24}(50\times 24+3(24)^2-\frac{1}{9}(24)^3)$

$=58^{\circ}$

Hence, the value of average temperature during that period= $58^{\circ}$

Answer from: Quest

i think it is everything has to equal 180 so, if i correct 4=c. 6=d. 8=a. 7=b. and the last question would be c. because they are parallel lines and they are congruent.

step-by-step explanation:

the reason everything has to equal to 180 is because that is how it works. i'm guessing that is its rule. 4 and 8 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 6 and 7 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 60+60 is 180. so, now you have your answer

Answer from: Quest

a. geometricb. g[h] = (3/4)·g[h-1]; g[1] = 384c. g[h] = 384·(3/4)^(h-1)d. g[12] ≈ 16.2183 gallonse. g[12] ≈ 64.8732 quarts

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt

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