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3 years ago

Examples of Trigonometry

Mathematics

2 answers:

Strike441 [17]3 years ago

8 0

Hope this picture helps you

frutty [35]3 years ago

4 0

Answer:Finding the length of a flagpole when you know 2 angles and the length of the shadow

Step-by-step explanation:

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Simply plz hurry thx

Lunna [17]

The answer to this question is D 17/24

8 0

3 years ago

Point B is between A and C.<br> AB=4x - 6, BC=2x+2, and AC=44 in.<br> Find AB and BC.

GREYUIT [131]

Please double check my work
AB = -2x
BC = 4x

7 0

4 years ago

What is 1000*90000 how will you solve it

exis [7]

Answer:

90,000,000

Step-by-step explanation:

If it is to hard for you, you can:

-Multiply the first digits by each other. 1 x 9 = 9.

-Then, add up all the zeros. In this case, there are, 7 zeros.

-Then, combine the 9 and add the seven zeros to make it:

90,000,000.

8 0

3 years ago

A 2-column table has 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2. The second column is la

andriy [413]

Answer:

Initial value is $\frac{1}{2}$ .

Step-by-step explanation:

Let us assume that the exponential function is $f(x) = a(b)^{x}$ ...... (1)

Now, the point $(0,\frac{1}{2})$ will satisfy the above equation.

Now, this point on the graph of the exponential function will be sufficient to find the initial value of the function (1) i.e. a.

Now, putting the value x = 0 and $y = \frac{1}{2}$ in the equation (1) we get, $f(x) = a(b)^{x}$

⇒ $\frac{1}{2} = a(b)^{0} = a$

⇒ $a = \frac{1}{2}$ (Answer)

10 0

3 years ago

Read 2 more answers

Find the complete time-domain solution y(t) for the rational algebraic output transform Y(s):_________

bija089 [108]

Answer:

y(t)= 11/3 e^(-t) - 5/2 e^(-2t) -1/6 e^(-4t)

Step-by-step explanation:

$Y(s)=\frac{s+3}{(s^2+3s+2)(s+4)} + \frac{s+3}{s^2+3s+2} +\frac{1}{s^2+3s+2}$

We know that $s^2+3s+2=(s+1)(s+2)$ , so we have

$Y(s)=\frac{s+3+(s+3)(s+4)+s+4}{(s+1)(s+2)(s+4)}$

By using the method of partial fraction we have:

$Y(s)=\frac{11}{3(s+1)} - \frac{5}{2(s+2)} -\frac{1}{6(s+4)}$

Now we have:

$y(t)=L^{-1}[Y(s)](t)$

Using linearity of inverse transform we get:

$y(t)=L^{-1}[\frac{11}{3(s+1)}](t) -L^{-1}[\frac{5}{2(s+2)}](t) -L^{-1}[\frac{1}{6(s+4)}](t)$

Using the inverse transforms

$L^{-1}[c\frac{1}{s-a}]=ce^{at}$

we have:

$y(t)=11/3 e^{-t} - 5/2 e^{-2t} -1/6 e^(-4t)$

6 0

4 years ago