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Evgesh-ka [11]
3 years ago
14

URGENT 30 POINTS An engineer needs to find the length of a beam to support the beam holding the wall. If the two beams are perpe

ndicular to one another, and a=4 m and b=1 m, find z. Give your answer to one decimal place.

Mathematics
1 answer:
alexandr402 [8]3 years ago
5 0
The answer is not m= 95.
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Setler [38]

Answer:

x>3

Step-by-step explanation:

7 0
2 years ago
Write each expression as an algebraic​ (nontrigonometric) expression in​ u, u > 0.
max2010maxim [7]

Answer:

\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0

Step-by-step explanation:

We want to write the trignometric expression:

\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)\text{ where } u>0

As an algebraic equation.

First, we can focus on the inner expression. Let θ equal the expression:

\displaystyle \theta=\sec^{-1}\left(\frac{u}{10}\right)

Take the secant of both sides:

\displaystyle \sec(\theta)=\frac{u}{10}

Since secant is the ratio of the hypotenuse side to the adjacent side, this means that the opposite side is:

\displaystyle o=\sqrt{u^2-10^2}=\sqrt{u^2-100}

By substitutition:

\displaystyle= \sin(2\theta)

Using an double-angle identity:

=2\sin(\theta)\cos(\theta)

We know that the opposite side is √(u² -100), the adjacent side is 10, and the hypotenuse is u. Therefore:

\displaystyle =2\left(\frac{\sqrt{u^2-100}}{u}\right)\left(\frac{10}{u}\right)

Simplify. Therefore:

\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0

4 0
2 years ago
Help me please please
yanalaym [24]

Answer:

Hold on a sec i cant remember

3 0
2 years ago
Divide. Write the quotient in lowest terms. 3 3/8 ÷ 9
Troyanec [42]
3 3/8 = 27/8 
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7 0
3 years ago
A scientist calculated the mean and standard deviation of a data set to be = 120 and 0-9. She then found that she was
Lapatulllka [165]

Answer:

The awnser on edge is C)147

Step-by-step explanation:

Hope this helped

5 0
2 years ago
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