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bogdanovich [222]

2 years ago

8

A person whose eye level is 20 feet above the ground is looking at a building. The angle of elevation to the top of the building

is 31º and the angle of depression to the bottom of the building is 24∘. What is the height (in feet) of the building? Round your answer to the nearest foot.

1 answer:

Inessa [10]2 years ago

8 0

Distance of the person from the building (x)

tan 24 = 20/x => x = 20/tan 24 = 44.92 ft

The height of the building above the height of the person and the distance of the person from the building forms two legs of right angled triangle.

Therefore, if h is the height of the building, then;
tan 31 = (h-20)/44.92
h-20 = 44.92 tan 31
h = (44.92 tan 31) + 20 = 26.99 + 20 = 46.99 ft

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12) Write the equation of a rational function

aev [14]

Answer:

$\frac{4(x - 5)(x + 7)(x - 12)}{(x + 1)(x)(x - 12)}$

Step-by-step explanation:

A rational function is

$\frac{p(x)}{q(x)}$

where q(x) doesn't equal zero.

If p is a asymptote, or hole at that value, then we will use

$(x - p)$

Step 1: We have asymptote as 0 and -1 so our denomiator will include

$(x - 0)(x - ( - 1)$

Which is

$(x)(x + 1)$

So our denomator so far is

$\frac{p(x)}{x(x + 1)}$

Step 2: Find Holes.

Since 12 is the value of the hole,

$(x - 12)$

is a the binomial.

This will be both on the numerator and denomator so qe have

$\frac{(x - 12)}{x(x + 1)(x - 12)}$

Step 3: Put the x intercepts in the numerator.

Since 5 and -7 is the intercepts,

$\frac{(x - 12)(x - 5)(x + 7)}{x(x + 1)(x - 12)}$

Step 4: Horinzontal Asymptotes,

Multiply the numerator and denomiator out fully,

$\frac{ {x}^{3} - 10 {x}^{2} - 59x + 420 }{ {x}^{3} - 12 {x}^{2} + x - 12}$

Take a L

look at the coefficients,

Notice they have the same degree,3, this means if we divide the leading coefficents, we will get our horinzonral asymptote.

Multiply the numerator by 4.

$\frac{4(x - 12)(x - 5)(x - 7)}{x(x + 1)(x - 12)}$

Above is the function,

5 0

2 years ago

Cornelia cut equal lengths of ribbon each 7/10 ft long. The ribbon was 3 1/2 ft long.How many pieces did Cornelia cut?

Ilya [14]

she cut two and two thirds 2 2/3

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2 years ago

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Nikitich [7]

The answer to this is B

4 0

3 years ago

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- 7 - w <10 what is the answer

sesenic [268]

-7-w<10
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Hope this helps :)

3 0

2 years ago

Uninhibited growth can be modeled by exponential functions other than A(t) =Upper A 0 e Superscript kt. For example, if an i

laila [671]

The question is incomplete. Here is the complete question.

Uninhibited growth can be modeled by exponential functions other than $A(t)=A_{0}e^{kt}$ . for example, if an initial population P₀ requires n units of time to triple, then the function $P(t)=P_{0}(3)^{\frac{t}{n} }$ models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.

a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form $P(t)=P_{0}(3)^{\frac{t}{n} }$ that models the population.

b) What will the population be in 47 days?

c) When wil the population reach 750?

d) Express the model from part (a) in the form $A(t)=A_{0}e^{kt}$ .

Answer: a) $P(t)=50(3)^{\frac{t}{30} }$

b) P(t) = 280 insects

c) t = 74 days

d) $A(t)=50e^{0.037t}$

Step-by-step explanation:

a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is

$P(t)=50(3)^{\frac{t}{30} }$

b) In t = 47 days:

$P(t)=50(3)^{\frac{t}{30} }$

$P(47)=50(3)^{\frac{47}{30} }$

$P(47)=50(3)^{1.567}$

P(47) = 280

In 47 days, population of insects will be 280

c) P(t) = 750

$750=50(3)^{\frac{t}{30} }$

$\frac{750}{50}=(3)^{\frac{t}{30} }$

$(3)^{\frac{t}{n} }=15$

Using the property <u>Power</u> <u>Rule</u> of logarithm:

$log(3)^{\frac{t}{30} }=log15$

$\frac{t}{30}log(3)=log15$

$t=\frac{log15}{log3} .30$

t = 74

To reach a population of 750 insects, it will take 74 days

d) To express the population growth into the described form, determine the constant k, using the following:

A(t) = 3A₀ and t = 30

$A(t)=A_{0}e^{kt}$

$3A_{0}=A_{0}e^{30k}$

$3=e^{30k}$

Use Power Rule again:

$ln3=ln(e^{30k})$

$ln3=30k$

$k=\frac{ln3}{30}$

k = 0.037

Equation for exponential growth will be:

$A(t)=50e^{0.037t}$

3 0

2 years ago