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4 years ago

7

Kyle has a notebook for each of his 5 classes. He puts 6 stickers on each notebooks.there are 10 stickers on each sheet.how many

sheets will Kyle use.

1 answer:

monitta4 years ago

4 0

5x6=30... 30/10=3.
He will use 3 sheets.

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The sum of three consecutive integers is 114. what are the three integers?

RSB [31]

Answer: 37, 38, and 39

Step-by-step explanation: This problem states that the sum of 3 consecutive integers is 114 and it asks us to find the integers.

3 consecutive integers can be represented as followed.

X ⇒ <em>first integer</em>

X + 1 ⇒ <em>second integer</em>

X + 2 ⇒ <em>third integer</em>

<em />

Since the sum of our 3 consecutive integers is 114, we can set up an equation to represent this.

X + X + 1 + X + 2 = 114

We can simplify on the left side by combining the X's and the numbers.

3x + 3 = 114

-3 -3 ← <em>subtract 3 from both sides of the equation</em>

3x = 111

÷3 ÷3 ← <em>divide both sides of the equation by 3</em>

<em> </em>X = 37

X ⇒ <em>first integer = 37</em>

X + 1 ⇒ <em>second integer = 38</em>

X + 2 ⇒ <em>third integer = 39</em>

<em />

<em>Therefore, our 3 consecutive integers are 37, 38, and 39.</em>

5 0

4 years ago

Read 2 more answers

What is the sin, cos, and tan of -495 degrees, and -20π/3 ??

Diano4ka-milaya [45]

Sin-495 = 0.98
cos-495 = 0.19
tan-495 = 4.95

sin<span>-20π/3</span> = 0
cos<span>-20π/3</span> = 0.3
tan<span>-20<span>π/3</span></span> = 0

7 0

3 years ago

Use the Binomial Theorem to expand the binomial.

Artist 52 [7]

<h3><u>d^3 - 4bd^2 + 16b^2d - 64d^3 is the expanded binomial.</u></h3>

The binomial theorem involves Pascal's triangle, and essentially gives you the coefficients for the formula you're going to use to expand it.

In this case, the coefficients will be 1, 3, 3, and 1.

We can set up our formula like this:

(a + (-b)) = a^3 + a^2b + ab^2 + b^3

Now we can just plug in our values:

(d + (-4b))^3 = d^3 + d^2(-4b) + d(-4b)^2 + (-4b)^3

Now, we can simplify the equation.

(d + (-4b))^3 = d^3 - 4bd^2 + 16b^2d - 64d^3

4 0

3 years ago

Read 2 more answers

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

I need help with this

Wewaii [24]

Answer:

12x+24x+18 and 36x+24-6 can both be simplified into 36x+18, meaning the equation is an identity

3 0

3 years ago