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Bas_tet [7]
3 years ago
5

I will give brainliest to whoever is correct.

Mathematics
2 answers:
disa [49]3 years ago
7 0
Im going to say C hopefully I’m right
Dovator [93]3 years ago
5 0

Answer:

c

Step-by-step explanation:

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What is the length of AD?
Setler [38]
We find the length of AD using the similarity of right triangle property. See image attached.

AD² = BD × CD
AD² = 9 × 36
AD² = 324
AD² = 18²
AD = 18

The length of AD is 18 units

7 0
3 years ago
Joel can either work 2 days for 11 hours each day or 4 days for 6 hours each day. He takes the bus for $2.25 each way. On days h
Oxana [17]
Joel's work makes $9.50/ hr. During the 2 Day, 11 hrs shift, he earns a total of $217.55 but spends a cash of $9 on his bus. Therefore, the money left is $208.55. While on his 4 day, 6 hrs shift, he earns a total of $228 but spends $18, with a total money of $210. The option that he should take to make the most money is the 4 days for 6 hrs shift. 
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3 years ago
How can I this math equation -5 -x =-8
Liula [17]

Answer: x = 3

Step-by-step explanation:

-5 -x = -8

-× = -8 + 5

-x = -3

x = 3

6 0
3 years ago
PLEASE HELP !! ILL GIVE BRAINLIEST *EXTRA 40 POINTS* DONT SKIP :(( .!
Charra [1.4K]

Answer: no solution

Step-by-step explanation:

They have the same slope which means they are parallel lines that will never cross or intersect. This mean that there is no solution

7 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
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