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2 years ago

15

What is 8.05 × 10-5 in standard notation?

1 answer:

Makovka662 [10]2 years ago

8 0

Answer:0.0000805

Step-by-step explanation:

Edgenuity says it’s right

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1.17 − 0.07a + (−3.92a) What is this ????

umka21 [38]

Answer: 17-3.99a

Step-by-step explanation:

Remove parentheses.

17- 0.07a- 3.92a17−0.07a− 3.92a

Collect like terms.

17+(−0.07a−3.92a)

Simplify.

17−3.99a

7 0

2 years ago

PLEASE HELP THIS A QUIZ DUE AT 9:00!

Svetllana [295]

Answer:

mmmm 7$ per week

Step-by-step explanation:

just divide 28/4=7

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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 $y_2(t)=t^{-5}$ .

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3 years ago

Solve each equation by graphing the related function. if the equation has no real number, solution, write no solution 144-p2=0

Molodets [167]

P= +or- 12. Because i did the maths

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3 years ago

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0.1% of what number is 4?

finlep [7]

$0.4$
that is my answer

4 0

3 years ago

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