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3 years ago

12

WILL MARK BRAINLIEST A stamp collector bought 7 rare stamps for her collection. Each stamp cost $1,953. How much did the collect

or spend on the rare stamps?*

1 answer:

IceJOKER [234]3 years ago

8 0

Answer:

$13,671

Step-by-step explanation:

Each stamp cost $1,953, and the collector bought 7 of them.

1953 * 7 = 13671

The collector spent $13,671 on the stamps.

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What is the major difference between Fourier Transform and Z-Transform??.

never [62]

<span>The Z transform is related to Laplace transform and is applied to analyze digital circuits. The Fourier transform, on the other hand, is a type z-transform, i.e z-transform which is used in spectrum analysis and calculation of the energy density of the signal. The difference lies on the field of application.</span>

7 0

3 years ago

Needing help ASAP, use integers

damaskus [11]

The expression to answer the question is 785 - 570 + 595.

785 - 570 + 595 = 810

Hopefully I could help. ;)

5 0

3 years ago

The curves r1(t) = 2t, t2, t4 and r2(t) = sin t, sin 5t, 2t intersect at the origin. Find their angle of intersection, θ, correc

masya89 [10]

Answer:

Therefore the angle of intersection is $\theta =79.48^\circ$

Step-by-step explanation:

Angle at the intersection point of two carve is the angle of the tangents at that point.

Given,

$r_1(t)=(2t,t^2,t^4)$

and $r_2(t)=(sin t , sin5t, 2t)$

To find the tangent of a carve , we have to differentiate the carve.

$r'_1(t)=(2,2t,4t^3)$

The tangent at (0,0,0) is [ since the intersection point is (0,0,0)]

$r'_1(0)=(2,0,0)$ [ putting t= 0]

$|r'_1(0)|=\sqrt{2^2+0^2+0^2} =2$

Again,

$r'_2(t)=(cos t ,5 cos5t, 2)$

The tangent at (0,0,0) is

$r'_2(0)=(1 ,5, 2)$ [ putting t= 0]

$|r'_1(0)|=\sqrt{1^2+5^2+2^2} =\sqrt{30}$

If θ is angle between tangent, then

$cos \theta =\frac{r'_1(0).r'_2(0)}{|r'_1(0)|.|r'_2(0)|}$

$\Rightarrow cos \theta =\frac{(2,0,0).(1,5,2)}{2.\sqrt{30} }$

$\Rightarrow cos \theta =\frac{2}{2\sqrt{30} }$

$\Rightarrow cos \theta =\frac{1}{\sqrt{30} }$

$\Rightarrow \theta =cos^{-1}\frac{1}{\sqrt{30} }$

$\Rightarrow \theta =79.48^\circ$

Therefore the angle of intersection is $\theta =79.48^\circ$ .

8 0

2 years ago

Write the given differential equation in the form L(y) = g(x), where L is a linear differential operator with constant coefficie

melamori03 [73]

Answer:

The complete solution is

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

Step-by-step explanation:

Given differential equation is

3y"- 8y' - 3y =4

The trial solution is

$y = e^{mx}$

Differentiating with respect to x

$y'= me^{mx}$

Again differentiating with respect to x

$y''= m ^2 e^{mx}$

Putting the value of y, y' and y'' in left side of the differential equation

$3m^2e^{mx}-8m e^{mx}- 3e^{mx}=0$

$\Rightarrow 3m^2-8m-3=0$

The auxiliary equation is

$3m^2-8m-3=0$

$\Rightarrow 3m^2 -9m+m-3m=0$

$\Rightarrow 3m(m-3)+1(m-3)=0$

$\Rightarrow (3m+1)(m-3)=0$

$\Rightarrow m = 3, -\frac13$

The complementary function is

$y= Ae^{3x}+Be^{-\frac13 x}$

y''= D², y' = D

The given differential equation is

(3D²-8D-3D)y =4

⇒(3D+1)(D-3)y =4

Since the linear operation is

L(D) ≡ (3D+1)(D-3)

For particular integral

$y_p=\frac 1{(3D+1)(D-3)} .4$

$=4.\frac 1{(3D+1)(D-3)} .e^{0.x}$ [since $e^{0.x}=1$ ]

$=4\frac{1}{(3.0+1)(0-3)}$ [ replace D by 0 , since L(0)≠0]

$=-\frac43$

The complete solution is

y= C.F+P.I

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

4 0

2 years ago

Minchanka [31]

Answer:

B

Step-by-step explanation: edge

3 0

2 years ago