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2 years ago

11

Please help ill give BRAINLIEST

1 answer:

S_A_V [24]2 years ago

6 0

Answer:

B and D

Step-by-step explanation:

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My grade is a 77%, if i get a 0% on a paper worth 30% of my grade, what will my grade be now??

blsea [12.9K]

$Grade=Paper+Other$
$Grade=(Grade_{paper}\times Value_{paper})+(Grade_{other}\times Value_{other})$
$Grade=(0 \times 30\%)+(77\% \times (1-30\%))$
$Grade=(0 \times 30\%)+(77\% \times 70\%)$
$Grade=0+53.9\%$
$Grade=53.9\%$

6 0

3 years ago

Given the function g(x) = 4(3)x, Section A is from x = 1 to x = 2 and Section B is from x = 3 to x = 4.

LenaWriter [7]

Easy peasy

the average rate of change in section A is the slope from (1,g(1)) to (2,g(2))
the average rate of chagne in section B is the slope from (3,g(3)) to (4,g(4))

A.

section A
g(1)=4(3)^1=12
g(2)=4(3)^2=4(9)=36
slope=(36-12)/(2-1)=24/1=24

section B
g(3)=4(3)^3=4(27)=108
g(4)=4(3)^4=4(81)=324
slope=(324-108)/(4-3)=216/1=216

section A has an average rate of change of 24
section B has an average rate of change of 216

3 0

3 years ago

The diameter of a circle is 9 m. Find the circumference to the nearest tenth.

DerKrebs [107]

Answer:

Circumference = diameter × π

Take π =3. 14

9×3. 14 = 28. 26 ~ 28. 3

5 0

3 years ago

Lisa has a rectangular trampoline that is 6 feet long and 12 feet wide. What is the area of her trampoline in square feet?

Cloud [144]

6*12 = 72ft²

The area of the trampoline is 72ft²

Hope this helps! :)

3 0

3 years ago

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

3 years ago