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3 years ago

6. Lines C and D below are cut by transversal, t. Which of

Mathematics

1 answer:

ra1l [238]3 years ago

8 0

Answer:

m∠8 = m∠2, option G

Step-by-step explanation:

Angles 8 and 2 are alternate interior angles, which means they are equal. Thus, the answer is option G, m∠8 = m∠2

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What is the difference between a number a and 6 written in a expression

iVinArrow [24]

Answer:

x-6

Step-by-step explanation:

Let the number be x then the difference would be x-6

3 0

3 years ago

Ped

Stells [14]

So the correct statement is:

" The GCF of the numbers in each term in the expression is 2"

<h3></h3><h3>How to rewrite the expression as a product?</h3>

Here we have the expression:

12d - 26c

To rewrite ti as a product, we need to find the greatest common factor between 12 and 26.

The decomposition of these two numbers is:

12 = 2*2*3

26 = 2*13

The greatest common factor between the two numbers is 2, then we can rewrite:

12d - 26c = 2*(6d - 13c)

So the correct statement is:

" The GCF of the numbers in each term in the expression is 2"

If you want to learn more about greatest common factors:

brainly.com/question/219464

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6 0

2 years ago

I have to find the sides of the angles on this photo

4vir4ik [10]

Step-by-step explanation:

may be those can be QT and QN

8 0

3 years ago

Write the expression 12-2 in simplest form.

Vesnalui [34]

well, if the expression is simply 12-2, then you would do simple subtraction and your answer would be 10. that would be simplest form. you cannot take it any simpler because there is nothing else left now. your answer would be 10.

3 0

3 years 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 $N\in\mathbb N$ .

6 0

3 years ago