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2 years ago

Find the length of the altitude to the hypotenuse. round the answer to the nearest tenth, if needed.

Mathematics

1 answer:

Amiraneli [1.4K]2 years ago

8 0

Answer : The correct option is, (B) 56.8

Step-by-step explanation :

Right Triangle Altitude Theorem:

The measure of altitude drawn from the vertex of the right angle of a right angled triangle to its hypotenuse is the geometric mean between the measures of the two segments of the hypotenuse.

We are drawing a line from vertex to hypotenuse side.

As per theorem:

$\frac{BD}{AB}=\frac{CB}{AC}$

$\frac{BD}{60}=\frac{175}{185}$

$BD=\frac{175\times 60}{185}$

$BD=56.8$

Therefore, the length of the altitude to the hypotenuse is 56.8

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2 years ago

I will pick BRAINLIEST.

serg [7]

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7 0

2 years ago

Can someone help answer and explain to me how to solve this? log (lowercase 6) 1/36

MissTica

Answer: The log simplifies to -2

-------------------------------------------
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Explanation:

We will use the log rule that log(x^y) = y*log(x). Call this log rule 1. This log rule basically allows us to pull the exponent down.

Another log rule that we will use is $\log_x\left(x\right) = 1$ where x is any positive real number but x = 1 is NOT allowed. Call this log rule 2.

Because 36 = 6^2, this means that 1/36 = 6^(-2)

-------------

So,

$\log_6\left(\frac{1}{36}\right)=\log_6\left(\frac{1}{6^2}\right)$

$\log_6\left(\frac{1}{36}\right)=\log_6\left(6^{-2}\right)$

$\log_6\left(\frac{1}{36}\right)=-2*\log_6\left(6\right)$ Use log rule 1 (see above)

$\log_6\left(\frac{1}{36}\right)=-2*1$ Use log rule 2 (see above)

$\log_6\left(\frac{1}{36}\right)=-2$

This means that the given expression simplifies to -2

You can use a calculator to type in "log(1/36)/log(6)" without quotes and you should get -2 as the answer

6 0

3 years ago

Read 2 more answers

A force of 50 pounds acts on an object at an angle of 45 degrees. A second force of 75 pounds acts on the object at an angle of

Contact [7]

Answer:

The magnitude of the resultant force is:

$R=100.3 \:pound$

The direction is:

$\theta=1.2^{\circ}$

Step-by-step explanation:

Let's find the components of each vector is x and y-directions first.

<u>Sum of x-component vector forces.</u>

$F_{tot-x}=F_{1}cos(45)+F_{2}cos(30)$

$F_{tot-x}=50cos(45)+75cos(30)$

$F_{tot-x}=100.3 \: pound$

<u>Sum of y-component vector forces.</u>

$F_{tot-y}=-F_{1}sin(45)+F_{2}sin(30)$

$F_{tot-y}=-50sin(45)+75sin(30)$

$F_{tot-y}=2.1 \: pound$

The magnitude of the resultant force is:

$R=\sqrt{100.3^{2}+2.1^{2}}$

$R=100.3 \:pound$

The direction is:

$tan(\theta)=\frac{2.1}{100.3}$

$\theta=arctan(\frac{2.1}{100.3})$

$\theta=1.2^{\circ}$

I hope it helps you!

4 0

2 years ago

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2 years ago