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3 years ago

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Can someone help me with 31 and 33

1 answer:

LenaWriter [7]3 years ago

8 0

31 between is 32 then 33

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What is the value of the expression [(1+5)x2-5]x2

Stella [2.4K]

[6 x 2 - 5] x 2
= (12 - 5 ) x 2
= 7 x 2
=14

3 0

3 years ago

Read 2 more answers

Helppppppppppp plzzzz

olasank [31]

3.14*r^2

Hope it helps you figure them out

8 0

3 years ago

Find the measure of x in each figure

uysha [10]

Answer:

x = 20

Step-by-step explanation:

The two angle are a linear pair. Their measures has a sum of 180 deg.

2x + 140 = 180

2x = 40

x = 20

7 0

3 years ago

The E. coli bacteria has a volume of 6 µm3 . In optimal conditions, an E. coli bacteria will double about every 30 minutes. Unde

Zinaida [17]

Answer:

It is going to take around 19 hours and 15 minutes for a single bacteria to grow to fill a thimble with volume 1 cm3.

It will take 57.6 hours to for the volume to fill the entire earth

Step-by-step explanation:

We can say that the volume of the bacteria is a geometric sequence, and each time moment is an arithmetic sequence.

Each geometric sequence has the following format:

${a, ar, ar^{2}, ar^{3},...}$

In which:

a is the first term

r is the common ratio.

We can find any term of the sequence by the following equation:

$x_{n} = ar^{(n-1)}$

Each arithmetic sequence has the following format:

${a, a+d, a+2d,...}$

In which:

a is the first term

d is the difference between the terms.

We can find any term of the sequence by the following equation:

$x_{n} = a + d(n)$

How i am going to solve this problem.

We have the sequence that is the volume of the bacteria:

Obs: $1(um)^{3} = 10^{-18}m^{3}$

$V_{n} = {6*10^{-18}, 12*10^{-18},...}$

$a = 6*10^{-18}$

$r = 2$

And the following arithmetic sequence that are the time(in hours).

$T_{n} = {0,0.5,1,...}$

$a = 0$

$d = 0.5$

$T_{n} = 0.5n$

How long will it take for a single bacterium to grow to fill a thimble with volume 1 cm3 ?

I am going to find the value of n for which $V_{n} = 1cm^{3}$ , then i find the value at this position in the arithmetic sequence. So

$1cm^{3} = 10^{-6}m^{3}$

$V_{n} = 6*10^{-18}*(2^{(n-1)})$

$10^{-6} = 6*10^{-18}*(2^{(n-1)})$

$\frac{10^{-6}}{6*10^{-18}} = 2^{(n-1)})$

Obs: $a^{b-c} = \frac{a^{b}}{a^{c}}$

$\frac{10^{12}}{6} = \frac{2^{n}}{2}$

$2^{n} = \frac{10^{12}}{3}$

Now, we have to apply these following logarithim proprierties to find the value of n:

$log a^{n} = n log a$

$log(\frac{a}{b}) = log a - log b$

$log 10^{n} = n$

log 2 = 0.30

log 3 = 0.48

$log 2^{n} = log(\frac{10^{12}}{3})$

$n log 2 = log(10^{12}) - log 3$

$0.30n = 12 - 0.48$

$0.30n = 11.52$

$n = 38.4$

Lets find $T_{38}$ and $T_{39}$ in the arithmetic sequence.

$T_{n} = 0.5n$

$T_{38} = 0.5*38 = 19$

$T_{39} = 0.5*39 = 19.5$

It is going to take around 19 hours and 15 minutes for a single bacteria to grow to fill a thimble with volume 1 cm3.

How long will it take for the volume to fill the entire earth 1.08 × 108 km3 ?

$1 (km)^{3} = 10^{9}m^{3}$

So

$1.08*10^{8} km^{3} = 1.08*10^{17} m^{3} = 108*10^{15}m^{3}$

Lets solve the same way as the first question.

$V_{n} = 6*10^{-18}*(2^{(n-1)})$

$108*10^{15} = 6*10^{-18}*(2^{(n-1)})$

$\frac{108*10^{15}}{6*10^{-18}} = 2^{(n-1)})$

$18*10^{33} = \frac{2^{n}}{2}$

$2^{n} = 36 * 10^{33}$

Aside from the proprierties seen in the first exercise, we also have that

log a*b = log a + log b

$log (2^{n}) = log(36 * 10^{33})$

$n log 2 = log 36 + log 10^{33}$

$0.30n = 1.56 + 33$

$0.30n = 34.56$

$n = \frac{34.56}{0.30}$

$n = 115.2$

$T_{115} = 0.5*115.2 = 57.6$

It will take 57.6 hours to for the volume to fill the entire earth

4 0

3 years ago

Use Distributive property 32x6)+(32x4) and rewrite

OverLord2011 [107]

Ily cor the same time as a result of the

8 0

3 years ago