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3 years ago

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A women standing on a cliff at the edge of the ocean spots a raft. Her eye level is 65 feet above sea level and the angle of dep

ression is 9° . How far is the raft from the base of the cliff?

1 answer:

Aliun [14]3 years ago

4 0

Answer:

the answer is C

Step-by-step explanation:

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Use the graph to find the unit rate for renting a kayak.

Firlakuza [10]

Answer:

Unit rate: $15 per hour

ordered pair: (1 ,15)

Step-by-step explanation:

30 ÷ 2 = 15

45 ÷ 3 = 15

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they both have 15, which means the graph is proportional and the unit rate is $15 per hour

7 0

2 years ago

Angle∠efg and angle∠gfh are a linear pair, mangle∠efgequals=55nplus+2020, and mangle∠gfhequals=33nplus+1616. what are mangle∠e

mylen [45]

A linear pair is a pair of angles that form a line. When added, they will equal 180.

< EFG + < GFH = 180
5n + 20 + 3n + 16 = 180
8n + 36 = 180
8n = 180 - 36
8n = 144
n = 144/8
n = 18

< EFG = 5n + 20.....5(18) + 20 = 110 <==
< GFH = 3n + 16.....3(18) + 16 = 70 <==

3 0

3 years ago

Read 2 more answers

How do u graph y=-2x

Anit [1.1K]

Answer:

if x=1 y=-2x1

Step-by-step explanation:

3 0

3 years ago

Which equation can be used to find the solution of (1/2)^2x = 32?

stira [4]

-2x=5 is the right answer

6 0

3 years ago

A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, t

ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

$T(t)=T_a+(T_0-T_a)e^{-kt}$

where $T_a$ is the ambient temperature, $T_0$ is the initial temperature, $t$ is the time and $k$ is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say $T_0$

We know that the ambient temperature is $T_a=65$ , so

$T(t)=65+(T_0-65)e^{-kt}$

We also know that when $t=5 \:min$ the temperature is $T(5)=35$ and when $t=10 \:min$ the temperature is $T(10)=50$ which gives:

$T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}$

$T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}$

Rearranging,

$35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}$

$50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}$

If we divide these two equations we get

$\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}$

$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

so the beam started out at 5 °F.

6 0

3 years ago