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2 years ago

Right triangles abc and dbc with right angle c are given below. If cos(a)=15,ab=12 and cd=2, find the length of bd.

Mathematics

2 answers:

dmitriy555 [2]2 years ago

8 0

see the attached figure to better understand the problem

we have that

$cos(A)=\frac{1}{5} \\ AB=12\ units\\ CD=2\ units$

Step 1

Find the value of AC

we know that

in the right triangle ABC

$cos (A)=(AC/AB)\\AC=AB*cos(A)$

substitute the values in the formula

$AC=12*(1/5)\\ AC=2.4\ units$

Step 2

Find the value of BC

we know that

in the right triangle ABC

Applying the Pythagorean Theorem

$AB^{2} =AC^{2}+BC^{2}\\ BC^{2}=AB^{2} -AC^{2}$

substitute the values

$BC^{2}=12^{2} -2.4^{2}\\BC^{2}= 138.24\\ BC=11.76\ units$

Step 3

Find the value of BD

we know that

in the right triangle BCD

Applying the Pythagorean Theorem

$BD^{2} =DC^{2}+BC^{2}$

substitute the values

$BD^{2} =2^{2}+11.76^{2}$

$BD=11.93\ units$

therefore

the answer is

the length of BD is 11.93 units

lubasha [3.4K]2 years ago

4 0

Answer: 17

Step-by-step explanation:

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Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.Tiles

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By definition, the average rate of change is given by:

$AVR = \frac{f(x2)-f(x1)}{x2-x1}$

We evaluate each of the functions in the given interval.

We have then:

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Evaluating for x = 3:

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Then, the AVR is:

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$AVR = \frac{18+2}{3+2}$

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$AVR = 4$

For f (x) = 3x - 8:

Evaluating for x =4:

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Evaluating for x = 5:

$f (5) = 3 (5) - 8\\f (5) = 15 - 8\\f (5) = 7$

Then, the AVR is:

$AVR = \frac{7-4}{5-4}$

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$AVR = 3$

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Evaluating for x = -3:

$f (-3) = (-3) ^ 2 - 2 (-3)\\f (-3) = 9 + 6\\f (-3) = 15$

Evaluating for x = 4:

$f (4) = (4) ^ 2 - 2 (4)\\f (4) = 16 - 8\\f (4) = 8$

Then, the AVR is:

$AVR = \frac{8-15}{4-(-3)}$

$AVR = \frac{-7}{4+3}$

$AVR = \frac{-7}{7}$

$AVR = -1$

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Evaluating for x = -1:

$f (-1) = (-1) ^ 2 - 5\\f (-1) = 1 - 5\\f (-1) = - 4$

Evaluating for x = 1:

$f (1) = (1) ^ 2 - 5\\f (1) = 1 - 5\\f (1) = - 4$

Then, the AVR is:

$AVR = \frac{-4-(-4)}{1-(-1)}$

$AVR = \frac{-4+4}{1+1}$

$AVR = \frac{0}{2}$

$AVR = 0$

Answer:

from the greatest to the least value based on the average rate of change in the specified interval:

f(x) = x^2 + 3x interval: [-2, 3]

f(x) = 3x - 8 interval: [4, 5]

f(x) = x^2 - 5 interval: [-1, 1]

f(x) = x^2 - 2x interval: [-3, 4]

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Answer:

The only value that is in the domains of both functions is 0

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Step-by-step explanation:

As the original function is

$f(x) = \sqrt{x}$

Since, domain is the set of all possible input values that define the function, and range is the set of all possible output values for all possible domain values for which the function is defined.

The domain of $f(x) = \sqrt{x}$ will be [0, ∞)

The range of $f(x) = \sqrt{x}$ will be [0, ∞)

Please check the attached figure a for visualizing the graph of $f(x) = \sqrt{x}$ .

Impact of double transformation:

When the function $f(x) = \sqrt{x}$ is reflected across x-axis, the function becomes $y = -\sqrt{x}$ after first transformation

After the second transformation across y-axis, the function $y = -\sqrt{x}$ becomes $g(x) = -\sqrt{-x}$

For

$g(x) = -\sqrt{-x}$

$-x$ must be equal to or greater than zero for $g(x) = -\sqrt{-x}$ to be defined i.e. -x ≥ 0.

So,

-x ≥ 0 can be written as x≤ 0

So,

The domain of $g(x) = -\sqrt{-x}$ will be (∞, 0]
The range of $g(x) = -\sqrt{-x}$ will be (∞, 0]

Please check the attached figure a for visualizing the graph of $g(x) = -\sqrt{-x}$ .

So, from the above discussion, we can say that

0 is the only that is in the domain of both function.
The range of g(x) is all values less than or equal to 0

So,

Only two statements are true about the functions f(x) and g(x) are true which are:

The only value that is in the domains of both functions is 0

The range of g(x) is all values less than or equal to 0

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