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3 years ago

12

How much deeper is the deepest canyon on mars than the deepest canyon on venus

2 answers:

djverab [1.8K]3 years ago

6 0

I googled it and the answer is 1,800 miles

azamat3 years ago

5 0

The grand canyon is 1,800 miles deeper the deepest canyon on Venus

Please mark as brianliest answer

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Two walkers 16 miles apart travel toward each other. One travels 3 miles per hour, and the other 4 miles per hour. If the latter

lapo4ka [179]

Answer:

28

For This Question

if you need steps ask me

7 0

2 years ago

Simplify -2.8f+0.9f-12-4

bogdanovich [222]

Answer:

-1.9f - 16

Step-by-step explanation:

-2.8f + 0.9f - 12 - 4 =
(-2.8 + 0.9)f - (12 + 4) =
-1.9f - 16

3 0

2 years ago

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Anyone mind helping?

mixas84 [53]

No solution would be your answer

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3 years ago

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Joanna bought 1 1/2 pounds of apples, 3/4 pound of strawberries, 1/2 pound of grapes, 1 3/4pounds of oranges, and 1 1/4 pounds o

Allushta [10]

The correct answer is 5.75 (5 3/4) pounds of fruit
1 1/2 +3/4 +1/2 + 1 3/4+ 1 1/4= 5 3/4

3 0

3 years ago

Read 2 more answers

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

Stells [14]

$y'=(t+y)^2-1$

Substitute $u=t+y$ , so that $u'=y'$ , and

$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

3 0

3 years ago