All categories

3 years ago

Please Help Quickly!!

1 answer:

3 0

To determine the extent of how many times greater is the number of bacteria in petri dish A than petri dish B, we simply have to take the ratio of the two.

1.75 x 10^18 / 6.25 x 10^15 = 280

So the bacteria in petri dish A is 280 times larger.

In scientific notation:

2.8 x 10^2

Hope this helps :)

You might be interested in

Tom predicted that the Giants would score three 3-point field goals, score two 6-point touchdowns, and score 1 extra point. If T

Tom [10]

3x3 = 9

2x6 = 12

12+9 = 21+1 =22

score would be 22 points

6 0

2 years ago

Read 2 more answers

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e

Alexxx [7]

Answer:

the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

Step-by-step explanation:

We are given the following information:

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in $\mu g/mL$

$C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})$

Thus, we are given the time interval [0,12] for t.

We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.

First, we differentiate C(t) with respect to t, to get,

$\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})$

Equating the first derivative to zero, we get,

$\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0$

Solving, we get,

$8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2$

At t = 0

$C(0) = 8(e^{(0)}-e^{(0)}) = 0$

At t = 2

$C(2) = 8(e^{(-0.8)}-e^{(-1.2)}) = 1.185$

At t = 12

$C(12) = 8(e^{(-4.8)}-e^{(-7.2)}) = 0.059$

Thus, the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

4 0

2 years ago

Can someone explain how to do #3?

dsp73

$\displaysyle (\sqrt 2)^{3}=(2^{\frac{1}{2}})^3=2^{\frac{1}{2} \cdot 3}=2^{\frac{3}{2}}=\sqrt {2^3}=\sqrt 8$

√8 is between 2 and 3, because 2²=4<8, but 3²=9>8. Also, our value is closer to 3 than to 2, so it is more than 2.5 and we have C and D options left.

Among these two numbers we find the one which is closer to √8.

C. 27=√729 ⇒ 2.7=√7.29

D. 28=√784 ⇒ 2.8=√7.84

Hence our answer is D) 2.8

5 0

3 years ago

Read 2 more answers

A cement walkway of uniform width has been built around an in-ground rectangular pool. the area of the walkway is 1344 square fe

Elza [17]

Answer:

6 feet.

Step-by-step explanation:

Dimensions of the Pool =80 feet long by 20 feet wide

Area of the walkway =1344 square feet.

If the width of the walkway=w

Length of the Larger Rectangle =80+2w
Width of the Larger Rectangle =20+2w

Area of the Walkway =Area of the Larger Rectangle - Area of the Pool

$1344=(80+2w)(20+2w)-(80*20)\\1344=80*20+160w+40w+4w^2-80*20\\4w^2+200w=1344\\4w^2+200w-1344=0\\We factorize\\4(w^2+50w-336)=0\\4(w^2-6w+56w-336)=0\\w(w-6)+56(w-6)=0\\(w-6)(w+56)=0\\w-6=0$ or $ w+56=0\\w=6$ ft or $ w=-56$ \\Therefore, the width of the walkway , w=6 feet.$