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3 years ago

13

EXPERTS/ACE ( and people that wanna actually help)

1 answer:

goblinko [34]3 years ago

4 0

Consistent means it has at least one solution.....inconsistent means there is no solution....coincident means it is the same line and there are infinite solutions.

ur 2 equations are the same line...therefore, ur answer is consistent and coincident

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PLEASE HELP!! 30 POINTS

Olenka [21]

Answer:

subject?

Step-by-step explanation:

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2 years ago

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A kitten’s mass at birth was 0.09 kilogram. The kitten gained approximately 0.084 kilogram each week. After how many weeks is th

Naddika [18.5K]

A kitten’s mass at birth was 0.09 kilogram. The kitten gained approximately 0.084 kilogram each week. After how many weeks is the kitten’s mass 1.098 kilograms?

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2 years ago

For a ride on a rental scooter, John paid an $8 fee to start the scooter plus 6 cents per minute of the ride. The total bill for

enot [183]

The linear equation is $8+0.06x=19.94$ and John rode the scooter for $199$ minutes.

Fee to start the scooter $=$ $ $8$ .

Fee to ride per minute $= 6$ cents $=$ $ $0.06$ .

Total bill $=$ $ $19.94$ .

Let $x$ be the minutes for which he rides the scooter.

The linear equation is formed as:

$8+0.06x=19.94$

$0.06x=19.94-8$

$0.06x=11.94$

$x=199$ minutes.

So, John rode the scooter for $199$ minutes.

Learn more about linear equation here:

brainly.com/question/24794429?referrer=searchResults

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1 year ago

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Newton's law of cooling is:

Mnenie [13.5K]

Answer:

$t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr$

So it would takes approximately 6.9 hours to reach 32 F.

Step-by-step explanation:

For this case we have the following differential equationÑ

$\frac{du}{dt}= -k (u-T)$

We can reorder the expression like this:

$\frac{du}{u-T} = -k dt$

We can use the substitution $w = u-T$ and $dw =du$ so then we have:

$\frac{dw}{w} =-k dt$

IF we integrate both sides we got:

$ln |w| = -kt +C$

If we apply exponential in both sides we got:

$w = e^{-kt} *e^c$

And if we replace w = u-T we got:

$u(t)= T + C_1 e^{-kt}$

We can also express the solution in the following terms:

$u(t) = (T_i -T_{amb}) e^{kt} +T_{amb}$

For this case we know that $k =-0.15 hr$ since w ehave a cooloing, $T_{i}= 70 F, T_{amb}=11F$ , we have this model:

$u(t) = (70-11) e^{-0.15t} +11$

And if we want that the temperature would be 32F we can solve for t like this:

$32 = 59 e^{-0.15 t} +11$

$21=59 e^{-0.15 t}$

$\frac{21}{59} = e^{-0.15 t}$

If we apply natural logs on both sides we got:

$ln (\frac{21}{59}) =-0.15 t$

$t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr$

So it would takes approximately 6.9 hours to reach 32 F.

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3 years ago

6 47/75 as a percentage

almond37 [142]

Answer:

662.6%

Step-by-step explanation:

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2 years ago