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2 years ago

Definitions given.

Mathematics

1 answer:

Yuki888 [10]2 years ago

8 0

The pitch of the roof is given by the ratio of the rise to the run of the roof, which is the slope of the rafter

$The \ pitch \ of \ the \ roof \ is \ \mathbf {\dfrac{6}{13}}$

Please find attached the simple <u>diagram of the house</u> with all measurements labeled

The reason the above value for the pitch and the diagram are correct are as follows:

The given parameters are;

Width of the section of the house on which the gable of the roof is to be placed = 50 feet

Length of the over hang on the sides of the roof = 1 foot each

Total span of the roof = 52 feet

Horizontal length of each rafter, the run = 26 feet

The height of the ridge above the top of the frame, the rise = 12 feet

Required:

To find the pitch of the roof

To draw a diagram of the house showing measurements

Solution:

$Pitch = \dfrac{Rise}{Run}$

$Pitch \ (slope) \ of \ roof = \dfrac{12}{26} =\dfrac{6}{13}$

Please find attached the <u>diagram</u> of the house showing the measurements for the roof parameters and the section of the house where the gable roof is to be placed

Learn more about finding the slope of a line here:

brainly.com/question/18990889

brainly.com/question/17105217

brainly.com/question/20987659

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The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

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