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2 years ago

There are 131 students at a small school and 35 of them are freshmen. What fraction of the students are freshmen?

Mathematics

1 answer:

IrinaK [193]2 years ago

6 0

Answer:

35/131

Step-by-step explanation:

Since 35 out of 131 students are freshmen, the fraction would be:

35/131

I hope this helps! :)

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Conflict is A. B. C. D.

QveST [7]

Conflict in Literature is the opposition of two or more characters or forces.

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3 years ago

Adding and subtracting polynomials

prohojiy [21]

Answer:

10z+10z+2+2+8z-8=20z+4

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2 years ago

A quadratic function and an exponential function are graphed below. Which graph most likely represents the exponential function?

Cloud [144]

Answer:

Step-by-step explanation:

The correct answer is the second one. The function g(x) is increasing "faster" than f(x), and this is true of exponential functions in general. This should make sense if you're familiar with the concept of limits at infinity.

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3 years ago

First will be marked as brainliest

Schach [20]

Start with

$\begin{cases}x+\frac{6}{y} = 6\\3x-\frac{8}{y}=5\end{cases}$

Assuming $y\neq 0$ , multiply both equations by y:

$\begin{cases}xy+6 = 6y\\3xy-8=5y\end{cases}$

Multiply the first equation by 3 and subtract the two equations:

$\begin{cases}3xy+18 = 18y\\3xy-8=5y\end{cases}\implies 26=13y \iff y=2$

Substitute the value for y in one of the equations to deduce the value for x:

$x+\frac{6}{2} = 6\iff x+3 = 6 \iff x= 3$

As suggested, we take the LCM and the equations become

$\begin{cases}\frac{3(x+1)+2(y-1)}{6}=8\\\frac{2(x-1)+3(y+1)}{6}=9\end{cases}$

Multiply both equations by 6 and you have

$\begin{cases}3x+3+2y-2}=48\\2x-2+3y+3=54\end{cases} \iff \begin{cases}3x+2y=47\\2x+3y=53\end{cases}$

From here, you can solve the system as in point 1.

Cross multiplication is exactly the method we used in point 1 to solve the equation: you multiply both equations so that they have the same coefficient for one of the variables, and then add/subtract. For example, if you multiply the first equation by 2 and add the the two equations, you'll simplify the y variable and will be able to solve for x.

The sum and multiplication of rational numbers is rational. So, we only need to prove that $\sqrt{3}$ is irrational, because:

$5+2\sqrt{3}$ is the sum between 5, that is rational, and $2\sqrt{3}$ . So, if this sum is irrational, is must be because of $2\sqrt{3}$
$2\sqrt{3}$ is the product of 2, that is rational, and $\sqrt{3}$ . So, if this product is irrational, it must be because of $\sqrt{3}$

Here's the proof that $\sqrt{3}$ is irrational, by contradiction:

$\sqrt{3}=\dfrac{a}{b} \iff \dfrac{a^2}{b^2}=3 \iff a^2=3b^2 \iff a=3k\\9k^2=3b^2 \iff b^2=3k^2 \iff b = 3m$

Here's the comment: we start by assuming that we can write $\sqrt{3}$ as the ratio between a and b, two integers with no common factors. We square both sides, and find you that a^2 is a multiple of 3, so a itself must be a multiple of 3. If we write a as 3k, we find out that b^2 is also a multiple of 3, and so is b.

We started assuming that a and b had no common factors, but we found out that they are both multiple of 3, hence the contradicition.

Every integer can be a multiple of six, or be 1,2,3,4 or 5 units away from a multiple of 6. In fact, if a number is 6 units away from a multiple of six, it is the next multiple of 6, and the count restarts.

This means that we can write every integer in one of these forms:

$6q,\ \ 6q+1,\ \ 6q+2,\ \ 6q+3,\ \ 6q+4,\ \ 6q+5$

A multple of 6 is even, so if we sum another even number we'll get an even result:

$6q,\ \ 6q+2,\ \ 6q+4\text{ are even}$

The remaining cases are the sum of an even number (6q) and an odd one (1, 3 or 5):

$6q+1,\ \ 6q+3,\ \ 6q+5\text{ are odd}$

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3 years ago

What is the prime factor of 32

ipn [44]

We factor out primes from 32 to find it's prime factorization.

32=2*16=2*2*8=2*2*2*4=2*2*2*2*2. 2 is a prime number, so this cannot be reduced further. Since we have five twos and no other primes, the factorization is 2^5, and the only prime factor is 2.

5 0

2 years ago

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