Whats 19.8 to 1 significant figure

What is the value of x of a complementary angle that is 10 and 2x

Usimov [2.4K]

Complementary angles together form a 90 degree angle. I don’t understand the question to well, but if two angles are next to each other and one of them is ten degrees, the other angle would be 80 is Together they form a right angle. Does my explain make sense? Sorry I hope this helps

7 0

3 years ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$= e^{-t-3} \left \{ {{1 \ \ \ \ \ t>3} \atop {0 \ \ \ \ \ t$

= $e^{-(t-3)} u (t-3)$

Recall that:

$y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

Then

$y(t) = 7 -7e^{-t} +e^{-(t-3)} u (t-3)$

$y(t) = 7 -7e^{-t} +e^{-t} e^{-3} u (t-3)$

$\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

3 0

3 years ago

Can someone please help me with this question?

Genrish500 [490]

It’s numberrrrr 333333333333333

4 0

3 years ago

The quotient of a number and -7 decreased by 2, is 10.

klasskru [66]

Hello!

This expression can be broken down into 3 separate parts:

1. "the quotient of a number and (-7)"
2. "decreased by 2"
3. "is 10"

We'll begin with the first part. The word "quotient" implies the result of a division problem. Therefore, this part can be represented by the following expression (let "x" represent the unknown number):

$\frac{x}{(-7)}$

Now we'll move on to the second part. If something is decreased by a value of 2, we know that 2 is being subtracted. Therefore, this part can be represented by the following expression:

– 2

Now we'll move on to the third and final part. The phrase "is 10" implies that the operations preceding this part are equal to 10. Therefore, this part can be represented by the following expression:

= 10

Finally, put the three parts together to create the following equation:

$\frac{x}{(-7)}$ – 2 = 10

The equation has now been fully translated. If, however, you are required to find the unknown value (x), begin by adding 2 to both sides of the equation:

$\frac{x}{(-7)}$ = 12

Now multiply both sides by (-7):

x = (-84)

We have now proven that the unknown number is equal to (-84).

I hope this helps!

3 0

2 years ago

Use yes or no Can you please help me

Gemiola [76]

The answer Yes, yes, no

7 0

3 years ago