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Flura [38]
3 years ago
7

Is the following a function? (2,5),(-9,5), (1,8), (3,2), (-2,1)? A.yes B.no

Mathematics
2 answers:
ohaa [14]3 years ago
4 0

Answer:

Yes

Step-by-step explanation:

Phantasy [73]3 years ago
4 0
Yes sorry if wrong tho
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What is the equation of the line that has a slope of 0 and passes through the point (6,-8)
Fofino [41]

\bf (\stackrel{x_1}{6}~,~\stackrel{y_1}{-8})~\hspace{10em} slope = 0 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-8)=0(x-6)\implies y+8=0\implies y=-8

5 0
3 years ago
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What is the area of this irregular shape 28in 7in 20in 30in 25in
Alex777 [14]

Answer:

b1=1b2=1h=7b1=1b2=6h=2b1=6b2=9h=2

7 0
3 years ago
DUE IN 5 MINS PLEASE HELP
Maksim231197 [3]

Answer:

12

Step-by-step explanation:

you can pack one pack in 20 min

there are three 20 min in one hour

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5 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
What is the position<br> number of the term<br> 125 in the sequence<br> f(n) = 2n+3?
spin [16.1K]

Answer:

61st term in the sequence

Step-by-step explanation:

125 = 2n + 3

122 = 2n

n = 61

8 0
3 years ago
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