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Kazeer [188]
3 years ago
9

A ladder is leaning against a building and makes 32° angle with the ground. The top of the ladder reaches 20 feet up on the buil

ding.
What is the length of the ladder?
Mathematics
1 answer:
IrinaK [193]3 years ago
6 0

Answer:

37.74 ft

Step-by-step explanation:

An angle and its opposite side are known

Sin (of the angle)= opposite/ hypotenuse

The hypotenuse is equal to the length of the ladder, therefore solve the hypotenuse

Sin (32)= 20/ hypotenuse

hypotenuse= 20/ Sin (32)

Sin (32)=0.5299

hypotenuse= 20/ 0.5299

37.74 ft

Hopefully this helps :)

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Thermometer A shows the temperature in the morning. Thermometer B shows the temperature in the evening. What is the difference i
Evgesh-ka [11]

Answer:

(Thermometer B reading - Thermometer A reading)

Step-by-step explanation:

The thermometer reading aren't given in the question.

However, hypothetically.

The difference between two temperature values (morning and evening values) would be :

Temperature in the evening - morning temperature

Therefore,

If ;

Thermometer A reading = morning temperature

Thermometer B reading = evening temperature

Difference in the temperature :

(Thermometer B reading - Thermometer A reading)

3 0
3 years ago
If x < -3, then (x+31 = ?
mixer [17]

Answer:

answer≤27

Step-by-step explanation:

then the answer for x could be -4 or lower so -4+31 is 27 so the answer would be any number less than or equal to 27

6 0
3 years ago
Evaluate the following integral (Calculus 2) Please show step by step explanation!
Nuetrik [128]

Answer:

4\ln \left| \dfrac{1}{3}\sqrt{9+(\ln x)^2} + \dfrac{1}{3}\ln x \right|+\text{C}

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

\displaystyle \int \dfrac{4}{x\sqrt{9+(\ln(x))^2}}\:\:\text{d}x

Rewrite 9 as 3²:

\implies \displaystyle \int \dfrac{4}{x\sqrt{3^2+(\ln(x))^2}}\:\:\text{d}x

<u>Integration by substitution</u>

\boxed{\textsf{For }\sqrt{a^2+x^2} \textsf{ use the substitution }x=a \tan\theta}

\textsf{Let } \ln x=3 \tan \theta

\begin{aligned}\implies \sqrt{3^2+(\ln x)^2} & =\sqrt{3^2+(3 \tan\theta)^2}\\ & = \sqrt{9+9\tan^2 \theta}\\ & = \sqrt{9(1+\tan^2 \theta)}\\ & = \sqrt{9\sec^2 \theta}\\ & = 3 \sec\theta\end{aligned}

Find the derivative of ln x and rewrite it so that dx is on its own:

\implies \ln x=3 \tan \theta

\implies \dfrac{1}{x}\dfrac{\text{d}x}{\text{d}\theta}=3 \sec^2\theta

\implies \text{d}x=3x \sec^2\theta\:\:\text{d}\theta

<u>Substitute</u> everything into the original integral:

\begin{aligned} \implies \displaystyle \int \dfrac{4}{x\sqrt{9+(\ln(x))^2}}\:\:\text{d}x & = \int \dfrac{4}{3x \sec \theta} \cdot 3x \sec^2\theta\:\:\text{d}\theta\\\\ & = \int 4 \sec \theta \:\: \text{d}\theta\end{aligned}

Take out the constant:

\implies \displaystyle 4 \int \sec \theta\:\:\text{d}\theta

\boxed{\begin{minipage}{7 cm}\underline{Integrating $\sec kx$}\\\\$\displaystyle \int \sec kx\:\text{d}x=\dfrac{1}{k} \ln \left| \sec kx + \tan kx \right|\:\:(+\text{C})$\end{minipage}}

\implies 4\ln \left| \sec \theta + \tan \theta \right|+\text{C}

\textsf{Substitute back in } \tan\theta=\dfrac{1}{3}\ln x :

\implies 4\ln \left| \sec \theta + \dfrac{1}{3}\ln x \right|+\text{C}

\textsf{Substitute back in }  \sec\theta=\dfrac{1}{3}\sqrt{9+(\ln x)^2}:

\implies 4\ln \left| \dfrac{1}{3}\sqrt{9+(\ln x)^2} + \dfrac{1}{3}\ln x \right|+\text{C}

Learn more about integration by trigonometric substitution here:

brainly.com/question/28157322

brainly.com/question/28156093

8 0
2 years ago
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slamgirl [31]

Answer:

(A) 4.9\times 10^{-7}\ m

Step-by-step explanation:

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Speed of the wave (v) = 3.00\times 10^8\ m/s

The wavelength equation of the wave is given as:

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Here, \lambda \to wave\ wavelength

Now, plug in the given values and simplify.

\lambda=\frac{3\times 10^8}{6.1\times 10^{14}}\\\\\lambda=4.9\times 10^{-7}\ m

Therefore, the wavelength of the wave is option (A) 4.9\times 10^{-7}\ m

3 0
3 years ago
The value of the expression l-20l-l6l is
TiliK225 [7]
|-20|-|6|

Find the absolute value of |-20| and |6|. 

|-20| = 20
|6| = 6

|-20|-|6| becomes 20 - 6.

20 - 6 = 14

The value of the expression |-20|-|6| is 14.
4 0
3 years ago
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