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valentina_108 [34]

3 years ago

10

Through (-5, -4), slope = 7/5

1 answer:

Alexandra [31]3 years ago

7 0

Answer:

$y = \frac{7}{5} x + 3$

Step-by-step explanation:

Explained in picture

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Can someone find the slopes of these two please?

GenaCL600 [577]

Answer:

you put the points into the slope formula and solve : (y2-y1)/(x2-x1)

1. (4 -1) / (-2 -3)

3 / -5

2. (-1 -(-2)) / (5 - 0)

(-1 +2) / 5

1/ 5

Step-by-step explanation:

i hope this helped :)

8 0

3 years ago

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Write the following:<br>a) 16 as a fraction of 152<br>b) 15 as a fraction of 210<br>

Nesterboy [21]

Answer:

$\frac{16}{152}$

$\frac{15}{210}$

Step-by-step explanation:

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3 years ago

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Levart [38]

Answer:

1. 5200+5200=10400

2. 10400-361=10039

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8 0

3 years ago

Help Please :D!!!!!! v

Amiraneli [1.4K]

Answer:

A they are just reflected over the y axis

Step-by-step explanation:

5 0

3 years ago

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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