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3 years ago

Please I need help with this, I just need to know which ones are linear and which ones are not (1,2 and 3)

Mathematics

1 answer:

nirvana33 [79]3 years ago

8 0

Hello the answer is of course 4,5,6

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At 7:00pm one evening, the temperature was -8°F. At 6:00am the next morning me

OleMash [197]

Answer:

B mark me brainliest!!!!?

Step-by-step explanation:

It increased 37 F

A) -8-|25| is -8-25 which is -37 it didn’t decrease lol!

B) |-8-25| is |-37| which is 37 which is the answer

C) -37 nope

D) 8-25 is not 37 so the answer is B

4 0

3 years ago

Add [ –6 –2 2] + [–3 2 1].

skad [1K]

The answer is... A hope this helps

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3 years ago

(5, 3) and (6, r) m = -1

Rzqust [24]

Answer:

r = 2

Step-by-step explanation:

(2-3)/(6-5)

= -1/1

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3 years ago

Which expression is the best estimate of 1/5 times 24.5

butalik [34]

Answer:

Step-by-step explanation:

1.2 x 3.7 nearest tenth

1 x 4 nearest whole number

8 0

3 years ago

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of P lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for S.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for C :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick P such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

5 0

3 years ago