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3 years ago

14

The shadow of a vertical tower is 58 feet long when the angle of elevation of the sun is 36 degrees. What is the height of the t

ower to the nearest foot?

1 answer:

monitta3 years ago

6 0

42 feet. You have side opposite and one adjacent to the angle therefore you use tan. Then just rearrange for x

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Stuck On this one too. Think I’m missing a step

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Answer:

The answer is B) 152

Step-by-step explanation:

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2 years ago

rachel bought one double pack of CDs for $20. this is $4 less than 3/4 the cost of a triple pack of CDs. what is the price of a

Sophie [7]

The price of a triple pack of Cds is $30.

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3 years ago

Which values are solutions to the inequality? 5r≤6r−8 Select each correct answer. r = 7 r = 8 r = 9 r = 10

Ratling [72]

Answer:

r = 8,9,10

Step-by-step explanation:

The given inequality is :

5r≤6r−8 ...(1)

We need to find the value of r.

Subtracting 5r to both sides of the inequality .

5r-5r≤6r-5r−8

0≤r−8

r ≥ 8

Hence, the values of r are 8,9,10.

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2 years ago

Read 2 more answers

Cate and Elena were playing a card game. The stack of cards in the middle had 24 cards in it to begin with. Cate added 8 cards t

Yuki888 [10]

You start with 24, there are 8 added, there are 12 taken away, then 9 are taken away. The equation looks like:
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2 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

$\frac{dA}{A} = k\ dt$

Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

i.e.

$t = 6\ A = 500000$

So:

$500000= 1000000e^{k*6}$

Divide both sides by 1000000

$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

Recall that:

$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

7 0

2 years ago