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2 years ago

Find the least common denominator of 7/20 and 3/28

Mathematics

1 answer:

Genrish500 [490]2 years ago

7 0

Answer:

140

Step-by-step explanation:

We have been given two fractions $\frac{7}{20}$ and $\frac{3}{28}$ . We are asked to find the least common denominator of both fractions.

To find the least common denominator of both fractions, we will find least common multiple of 20 and 28.

Prime factorization of 20: $2\times 2\times 5$

Prime factorization of 28: $2\times 2\times 7$

Least common multiple of 20 and 28 would be: $2\times 2\times 7\times 5=140$ .

Therefore, the least common denominator of both fractions would be 140.

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If x, y, and z are integers greater than 1, and (327)(510)(z) = (58)(914)(xy), then what is the value of x?

algol [13]

Answer:

Statements (1) and (2) TOGETHER are NOT sufficient.

Explanation:

As in the equation (327)(510)(z) = (58)(914)(xy) there are THREE variables in total i.e. "x", "y" and "z" hence minimum three equations are required to find out values of all variables. Hence,

If the given number of equations is equal to total variable used in any of the equation, values of all the variables can be find out otherwise there can be unlimited number of solutions.

So, value of "x" cannot be determined with the given data.

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2 years ago

Fiona has met her goal of exercising 150 minutes a week. She has kept it up for six months and is sure that it is now solidly pa

cricket20 [7]

Answer:

Step-by-step explanation:

You are always able to do better, and Fiona needs to reflect

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1 year ago

Using a number cube and this hat with 5 different-colored marbles, which simulation would help you answer this question? During

Arlecino [84]

Answer:

see the attachment

Step-by-step explanation:

We assume that the question is interested in the probability that a randomly chosen class is a Friday class with a lab experiment (2/15). That is somewhat different from the probability that a lab experiment is conducted on a Friday (2/3).

Based on our assumption, we want to create a simulation that includes a 1/5 chance of the day being a Friday, along with a 2/3 chance that the class has a lab experiment on whatever day it is.

That simulation can consist of choosing 1 of 5 differently-colored marbles, and rolling a 6-sided die with 2/3 of the numbers being designated as representing a lab-experiment day. (The marble must be replaced and the marbles stirred for the next trial.) For our purpose, we can designate the yellow marble as "Friday", and numbers greater than 2 as "lab-experiment".

The simulation of 70 different choices of a random class is shown in the attachment.

_____

Comment on the question

IMO, the use of 70 trials is coincidentally the same number as the first 70 days of school. The calendar is deterministic, so there will be exactly 14 Fridays in that period. If, in 70 draws, you get 16 yellow marbles, you cannot say, "the probability of a Friday is 16/70." You need to be very careful to properly state the question you're trying to answer.

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2 years ago

What is the slope of the line passing through the points (5, 3) and (5, -9)? *

Mars2501 [29]

Answer:

Slope is 0

Step-by-step explanation: x stood the same so therefore the slope is 0 or infinite

4 0

2 years ago

If x^2y-3x=y^3-3, then at the point (-1,2), (dy/dx)?

zavuch27 [327]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2866883

_______________

 dy
Find —— for an implicit function:
 dx

x²y – 3x = y³ – 3

First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:

$\mathsf{\dfrac{d}{dx}(x^2 y-3x)=\dfrac{d}{dx}(y^3-3)}\\\\\\
\mathsf{\dfrac{d}{dx}(x^2 y)-3\,\dfrac{d}{dx}(x)=\dfrac{d}{dx}(y^3)-\dfrac{d}{dx}(3)}$

Applying the product rule for the first term at the left-hand side:

$\mathsf{\left[\dfrac{d}{dx}(x^2)\cdot y+x^2\cdot \dfrac{d}{dx}(y)\right]-3\cdot 1=3y^2\cdot \dfrac{dy}{dx}-0}\\\\\\
\mathsf{\left[2x\cdot y+x^2\cdot \dfrac{dy}{dx}\right]-3=3y^2\cdot \dfrac{dy}{dx}}$

 dy
Now, isolate —— in the equation above:
 dx

$\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3=3y^2\cdot \dfrac{dy}{dx}}\\\\\\
\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3-3y^2\cdot \dfrac{dy}{dx}=0}\\\\\\
\mathsf{x^2\cdot \dfrac{dy}{dx}-3y^2\cdot \dfrac{dy}{dx}=-\,2xy+3}\\\\\\
\mathsf{(x^2-3y^2)\cdot \dfrac{dy}{dx}=-\,2xy+3}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{-\,2xy+3}{x^2-3y^2}\qquad\quad for~~x^2-3y^2\ne 0}$

Compute the derivative value at the point (– 1, 2):

x = – 1 and y = 2

$\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{-\,2\cdot (-1)\cdot 2+3}{(-1)^2-3\cdot 2^2}}\\\\\\
\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{4+3}{1-12}}\\\\\\
\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{7}{-11}}\\\\\\\\ \therefore~~\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=-\,\dfrac{7}{11}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: implicit function derivative implicit differentiation chain product rule differential integral calculus

6 0

2 years ago