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3 years ago

Can people help? When a grizzly bear hibernates, its heart rate drops to

Mathematics

1 answer:

avanturin [10]3 years ago

8 0

Like you said: 10 beats per minute, which is 20% percent of its normal value.

So: 10 per minute is 20% of the normal rate. You will divide each side by 0.2, which will result in 50 per min

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PLZ HELP TEST IS BEING TIMED!!! What are the slope and the y-intercept of the linear function that is represented by the equatio

Oksana_A [137]

Answer:

The slope is -10, and the y-intercept is 1

Step-by-step explanation:

The slope is in the equation and always has a variable next to it (In this case, -10x, so -10 is the slope)

The y-intercept is 1 and that is also in the equation and does not have any variables next to it (In this case, + 1, so it is 1)

3 0

3 years ago

Read 2 more answers

It takes james 2.5 minutes to type 150 words at that rate how mant words can james type in 6 minutes

sergey [27]

The number of words james can type in 6 minutes at the same rate is 360 words

Given:

Number of minutes to type 150 words = 2.5 minutes

let

number of words James can type in 6 minutes = x

Equate the ratio of the number of words to number of minutes

150 : 2.5 = x : 6

150/2.5 = x/6

cross product

150 × 6 = 2.5 × x

900 = 2.5x

x = 900/2.5

x = 360 words

Therefore, the number of words james can type in 6 minutes is 360 words

Learn more about ratio:

brainly.com/question/16981404

5 0

3 years 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.

7 0

4 years ago

60 POINTSSSS Answer correctly please this depeneds on my dumb GRADE PLEASE

Maru [420]

Answer:

Option 4

Step-by-step explanation:

It's a geometric progression.

After t years,

500(1.1^t)

After 6 years,

500(1.1⁶) = 885.7805

4 0

3 years ago

I need some help please

Naily [24]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers