Use common factors to write two fractions equivalent to 6/24

Mathematics

2 answers:

ale4655 [162]3 years ago

7 0

Answer: 3/12 or 12/48

Step-by-step explanation:

Flura [38]3 years ago

6 0

1/4 and 3/12 are equal to 6/24

divide 6 and 24 by 6 to get 1/4 and divide 6 and 24 by 2 to get 3/12

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Three-fourths of a number is greater than or equal to nine less than a number.

NARA [144]

Answer:

$\frac{3}{4} x \geqslant x - 9$

8 0

3 years ago

Will give Brainlest Answer please answer i need help

Rashid [163]

Answer:

C. {-1, 5, 8}

Step-by-step explanation:

Use each of the domain values in the function to see what the corresponding range value is.

f(-1) = 5 -3(-1) = 8

f(0) = 5 -3(0) = 5

f(2) = 5 -3(2) = -1

The range is the set of numbers {-1, 5, 8}.

_____

Additional comment

The values in a set are generally listed lowest to highest. The coefficient of x in the equation for f(x) is negative, meaning the lowest range value will correspond to the highest domain value. If you start by finding f(2) = -1, you immediately eliminate all answer choices except B and C.

Those choices differ only in the middle value, so you can tell which is correct by evaluating f(x) for the middle domain value: f(0) = 5. Only one answer choice has both -1 and 5 in the set.

(There are two answers here: how you work the problem, and how you game a multiple choice question.)

4 0

1 year ago

Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

Nat2105 [25]

By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .