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3 years ago

HELP ME PLEZ Im timed lol

Mathematics

2 answers:

goldenfox [79]3 years ago

8 0

Answer:

1.) $1

2.)$4

3.)$12

4.)3/4

Step-by-step explanation:

32(.5)=16

she had 16 left after lunch. 16(.25)=4

16-4=12

1/2=2/4

2/4+1/4=3/4

Alex17521 [72]3 years ago

3 0

Answer:

1) lunch 16

2) coffee 4

3) 12

4) 3/4ths or .75 or 75%

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Read 2 more answers

A biologist observed that a certain bacterial colony obeys the population growth law and that the colony triples every 4 hours.

MrRa [10]

Answer:

a) $P(t) = 2e^{0.275t}$

b) 54.225 square centimeters.

c) 2.52 hours

Step-by-step explanation:

The population growth law is:

$P(t) = P_{0}e^{rt}$

In which P(t) is the population after t hours, $P_{0}$ is the initial population and r is the growth rate, as a decimal.

In this problem, we have that:

The colony occupied 2 square centimeters initially, so $P_{0} = 2$

The colony triples every 4 hours. So

$P(4) = 3P_{0} = 6$

(a) An expression for the size P(t) of the colony at any time t.

We have to find the value of r. We can do this by using the P(4) equation.

$P(t) = P_{0}e^{rt}$

$6 = 2e^{4r}$

$e^{4r} = 3$

Applying ln to both sides, we get:

$4r = 1.1$

$r = 0.275$

$P(t) = 2e^{0.275t}$

(b) The area occupied by the colony after 12 hours.

$P(t) = 2e^{0.275t}$

$P(12) = 2e^{0.275*12}$

$P(12) = 54.225$

(c) The doubling time for the colony?

t when $P(t) = 2P_{0} = 2*2 = 4$ .

$P(t) = 2e^{0.275t}$

$4 = 2e^{0.275t}$

$e^{0.275t} = 2$

Applying ln to both sides

$0.275t = 0.6931$

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3 years ago

If cosx=1/12 and sinx>0 find tan2x

Liono4ka [1.6K]

$cosx=\frac{1}{12}\\\\sin^2x+cos^2x=1\\\\sin^2x+\left(\frac{1}{12}\right)^2=1\\\\sin^2x+\frac{1}{144}=1\\\\sin^2x=1-\frac{1}{144}\\\\sin^2x=\frac{144}{144}-\frac{1}{144}\\\\sin^2x=\frac{143}{144}\\\\sinx=\sqrt\frac{143}{144}\\\\sinx=\frac{\sqrt{143}}{12}$

$tan2x=\frac{sin2x}{cos2x}\\\\sin2x=2sinxcosx;\ cos2x=cos^2x-sin^2x\\\\sin2x=2\cdot\frac{\sqrt{143}}{12}\cdot\frac{1}{12}=\frac{2\sqrt{143}}{144}\\\\cos2x=\left(\frac{1}{12}\right)^2-\left(\frac{\sqrt{143}}{12}\right)^2=\frac{1}{144}-\frac{143}{144}=-\frac{142}{144}\\\\tan2x=\frac{2\sqrt{143}}{144}:\left(-\frac{142}{144}\right)=-\frac{2\sqrt{143}}{144}\cdot\frac{144}{142}=-\frac{\sqrt{143}}{71}$

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3 years ago