All categories

3 years ago

Multiple choice please answer please

Mathematics

1 answer:

joja [24]3 years ago

6 0

Answer:

Step-by-step explanation:

Using the cosine ratio in the right triangle

cosA = $\frac{adjacent}{hypotenuse}$ = $\frac{AC}{AB}$ = $\frac{2}{3}$ , thus

A = $cos^{-1}$ ( $\frac{2}{3}$ ) ≈ 48° → A

You might be interested in

3^3 - 5 x 9 + 6 - 29

xxMikexx [17]

Answer:

-41

Step-by-step explanation:

3^3-5x9+6-29

calculate exponents 3^3=27

now write as; 27-5x9+6-29

now multiply and divide (left to right) 5x9=45

now rewrite as 27-45+6-29

now add and subtract(left to right) 27-45+6-29= -41

6 0

3 years ago

Read 2 more answers

Are they unique more than one or no triangle

Tanya [424]

The measures of the angles of a triangle add to 180.

45 + 45 + 90 = 180
The measures of these three angles do add to 180, ,so there is at least one triangle with these angle measures.
Using AA triangle similarity, any triangle with the same angle measures will be similar.

From the figure you already see two triangles with angles 45-45-90. There is an infinite number of triangles with those angle measures.

8 0

3 years ago

Solve the following differential equation using using characteristic equation using Laplace Transform i. ii y" +y sin 2t, y(0) 2

kifflom [539]

Answer:

The solution of the differential equation is $y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)$

Step-by-step explanation:

The differential equation is given by: y" + y = Sin(2t)

<u>i) Using characteristic equation:</u>

The characteristic equation method assumes that y(t)= $e^{rt}$ , where "r" is a constant.

We find the solution of the homogeneus differential equation:

y" + y = 0

$y'=re^{rt}$

$y"=r^{2}e^{rt}$

$r^{2}e^{rt}+e^{rt}=0$

$(r^{2}+1)e^{rt}=0$

As $e^{rt}$ could never be zero, the term (r²+1) must be zero:

(r²+1)=0

r=±i

The solution of the homogeneus differential equation is:

$y(t)_{h}=c_{1}e^{it}+c_{2}e^{-it}$

Using Euler's formula:

$y(t)_{h}=c_{1}[Sin(t)+iCos(t)]+c_{2}[Sin(t)-iCos(t)]$

$y(t)_{h}=(c_{1}+c_{2})Sin(t)+(c_{1}-c_{2})iCos(t)$

$y(t)_{h}=C_{1}Sin(t)+C_{2}Cos(t)$

The particular solution of the differential equation is given by:

$y(t)_{p}=ASin(2t)+BCos(2t)$

$y'(t)_{p}=2ACos(2t)-2BSin(2t)$

$y''(t)_{p}=-4ASin(2t)-4BCos(2t)$

So we use these derivatives in the differential equation:

$-4ASin(2t)-4BCos(2t)+ASin(2t)+BCos(2t)=Sin(2t)$

$-3ASin(2t)-3BCos(2t)=Sin(2t)$

As there is not a term for Cos(2t), B is equal to 0.

So the value A=-1/3

The solution is the sum of the particular function and the homogeneous function:

$y(t)= - \frac{1}{3} Sin(2t) + C_{1} Sin(t) + C_{2} Cos(t)$

Using the initial conditions we can check that C1=5/3 and C2=2

<u>ii) Using Laplace Transform:</u>

To solve the differential equation we use the Laplace transformation in both members:

ℒ[y" + y]=ℒ[Sin(2t)]

ℒ[y"]+ℒ[y]=ℒ[Sin(2t)]

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]-s·y(0)-y'(0)=s²·Y(s) -2s-1

ℒ[y]=Y(s)

ℒ[Sin(2t)]= $\frac{2}{(s^{2}+4)}$

We replace the previous data in the equation:

s²·Y(s) -2s-1+Y(s) = $\frac{2}{(s^{2}+4)}$

(s²+1)·Y(s)-2s-1= $\frac{2}{(s^{2}+4)}$

(s²+1)·Y(s)= $\frac{2}{(s^{2}+4)}+2s+1=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)}$

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

Y(s)= $\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}$

Using partial franction method:

$\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}=\frac{As+B}{s^{2}+4} +\frac{Cs+D}{s^{2}+1}$

2s^{3}+s^{2}+8s+6=(As+B)(s²+1)+(Cs+D)(s²+4)

2s^{3}+s^{2}+8s+6=s³(A+C)+s²(B+D)+s(A+4C)+(B+4D)

We solve the equation system:

A+C=2

B+D=1

A+4C=8

B+4D=6

The solutions are:

A=0 ; B= -2/3 ; C=2 ; D=5/3

So,

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

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

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[ $-\frac{1}{3} \frac{2}{s^{2}+4}$ ]-ℒ⁻¹[ $2\frac{s }{s^{2}+1}$ ]+ℒ⁻¹[ $\frac{5}{3}\frac{1}{s^{2}+1}$ ]

$y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)$

3 0

3 years ago

Expanded form.2(b + c) *

11Alexandr11 [23.1K]

Answer:

Step-by-step explanation:

2b + 2c

4 0

3 years ago

Read 2 more answers

Solve each trigonometric equation such that 0 ≤ x ≤ 2????. Give answers in exact form.

sveticcg [70]

Answer:

(a) $x=\frac{5\pi }{3}$

(b) $x=\frac{\pi }{8}$

Step-by-step explanation:

It is given that $0\leq x\leq 2\pi$

(a) We have given equation $2sinx+\sqrt{3}=0$

$sinx=\frac{-\sqrt{3}}{2}$

$x=sin^{-1}(-0.866)$

$x=\frac{-\pi }{3}=2\pi -\frac{\pi }{3}=\frac{5\pi }{3}$

(b) $tan(2x)=1$

$2x=tan^{-1}1$

$2x=\frac{\pi }{4}$

$x=\frac{\pi }{8}$

7 0

3 years ago