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3 years ago

6

What is the answer I need it urgently

1 answer:

nordsb [41]3 years ago

5 0

It’s is 0.............

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Cesar is coordinator planning the Home Run Derby. You want to select players who deserve to be

madam [21]

Answer:

Wait From What I See Aren't You Just Supposed To Put A Name And Date?

Step-by-step explanation:

3 0

3 years ago

Find the value of x and y in the parallelogram below.

sashaice [31]

Answer:

x = 18°

y = 6

Step-by-step explanation:

in a parallelogram:

Any two opposite sides are congruent

and any two opposites angles are congruent:

then

y + 4 = 10

and 3x = 54

then

y = 6

and x = 54/3 = 18

8 0

3 years ago

Read 2 more answers

Work out the length X.

ohaa [14]

Answer:

16.34

Step-by-step explanation:

Is pythagoreons Theron

256 + 11.25 = c squared

267.25 = c squared

16.34 equals approximately c

8 0

3 years ago

Read 2 more answers

In this triangle, side AT = 24. Please give the value for the sine, cosine, and tangent for this diagram (Which is the marked an

sashaice [31]

The value of the sine, cosine and tangent of the figure will be found as follows:
a] Sine
sin x=(opposite)/(hypotenuse)
opposite=7
hypotenuse=25
thus:
sin x= 7/25

b] Cosine
cos x=adjacent/hypotensue
adjacent=24
hypotenuse=25
cos x=24/25

Tangent
Tan x=opposite/adajcent
opposite=7
adjacent=24
thus
tan x=7/24

5 0

3 years ago

Use the method of reduction of order to find a second solution to t^2y' + 3ty' – 3y = 0, t> 0 Given yı(t) = t y2(t) = Preview

Anestetic [448]

Let $y_2(t)=tv(t)$ . Then

${y_2}'=tv'+v$

${y_2}''=tv''+2v'$

and substituting these into the ODE gives

$t^2(tv''+2v')+3t(tv'+v)-3tv=0$

$t^3v''+5t^2v'=0$

$tv''+5v'=0$

Let $u(t)=v'(t)$ , so that $u'(t)=v''(t)$ . Then the ODE is linear in $u$ , with

$tu'+5u=0$

Multiply both sides by $t^4$ , so that the left side can be condensed as the derivative of a product:

$t^5u'+5t^4u=(t^5u)'=0$

Integrating both sides and solving for $u(t)$ gives

$t^5u=C\implies u=Ct^{-5}$

Integrate again to solve for $v(t)$ :

$v=C_1t^{-6}+C_2$

and finally, solve for $y_2(t)$ by multiplying both sides by $t$ :

$tv=y_2=C_1t^{-5}+C_2t$

$y_1(t)=t$ already accounts for the $t$ term in this solution, so the other independent solution is $y_2(t)=t^{-5}$ .

6 0

4 years ago