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3 years ago

Tom and Sam shared equally one third of a chocolate bar. What fraction of the chocolate bar did each child get?

Mathematics

2 answers:

Savatey [412]3 years ago

5 0

1/3 each , hope it helps

GuDViN [60]3 years ago

3 0

If two people shared one section worth one third then you divide one there by two (1/3)/2 aka (1/2)*(1/3) and you'll get 1/6

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Marianne has 5/8 package of peas. She cooks 2/3 of those peas for 5 people. If each person is served an equal amount, what fract

creativ13 [48]

5/8 and 2/3 so...
8 times 3 is 24 so now put your fraction over twenty four
15/24=5/8
15/24 so fifteen out of twenty fourths of the package is what she has. She takes ten out of the fifteenths because fifteen divided by three is five then multiply by two is ten. Then divide by five and you get two.
15/24. 10-> 15/24 cause... 15 divided by 3 ( from the two thirds she cooked) is 5 then * by 2... 5*2=10

5 0

3 years ago

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

3 years ago

Plz, help ASAP !!!!!!!!

mr_godi [17]

Answer:

145 degrees.

Step-by-step explanation:

This line is 180 degrees. So if you take 180 and subtract 35, you should get 145. Hope this helped!

-Kirito

6 0

3 years ago

Explain how solving 4x < -16 is different from solving -4x< 16.

nalin [4]

Answer:

Dividing by 4 and -4

Step-by-step explanation:

Solving 4x < -16 is different from solving -4x< 16, because for 4x < -16, you are dividing by a positive 4 on both sides, but for -4x< 16, you are dividing by -4 on both sides.

3 0

1 year ago

At a movie theatre adult tickets cost $10 and child ticjets cost $6. Which equation can be used to find s, the number of dollars

timurjin [86]

Answer:

s = 10k + 30

Step-by-step explanation:

Cost of Adults ticket = $10 per adult

Cost of Ticket = $6 per child

Number of adults = k

Number of Children = 5

s = Total cost

Total cost = (Cost per adult * number of adults) + (Cost per child * Number of children)

s = ($10 * k) + ($6 * 5)

s = $10k + $30

s = 10k + 30

4 0

3 years ago