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seropon [69]
3 years ago
9

What is the length of side s of the square shown below?

Mathematics
1 answer:
Tomtit [17]3 years ago
7 0
Maybe take a picture of it?
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2. Determine if the expressions are equivalent by comparing the models.
Ksju [112]

Answer:

answer 4 y +rdcmeint or 67456

Step-by-step explanation:

8 0
2 years ago
A geologist in South America discovers a
diamong [38]

Answer:

This is a radioactive decay / half-life problem.

Initial amount of C 14 = 100% Present Amount = 57%

k = .0001  (that value should be negative)

Nt = No * e^ (k*t)

You need to solve that equation for "time" (equation attached)

time = natural log (Ending amount / Starting Amount) / k

time = natural log (57% / 100%) / -.0001

time = ln (.57) / -.0001

time = -.56211891815 / -.0001

time = 5,621.2 years    (age of the bird skeleton)

These problems are quite complicated but I think I know this pretty well.

Need to know more? Visit my website (it's on the graphic).

Step-by-step explanation:

7 0
3 years ago
use the explicit formula an = a1 + (n - 1) * d to find the 500th term of the sequence below. 24,30,36,42,48,... A. 3018 B. 3042
marishachu [46]

Answer:

Option A.

Step-by-step explanation:

The given sequence is  

24, 30, 36, 42, 48, ...

It is an AP. Here,  

First term = 24

Common difference = 30-24 = 6

The given explicit formula for nth term is

a_n=a_1+(n-1)d

where, a_1 is first term, d is common difference.

Substitute a_1=24, d=6 \text{ and }n=500 in the above formula.

a_{500}=24+(500-1)(6)

a_{500}=24+(499)(6)

a_{500}=24+2994

a_{500}=3018

The 500th term of the sequence is 3018.

Therefore, the correct option is A.

3 0
2 years ago
Suppose that a teacher plans to give four students a quiz. The minimum possible score on the quiz is 0, and the maximum possible
denis23 [38]

Answer:

5 is the largest it could possibly be

8 0
3 years ago
tom is trying to write 3/47 as a decimal. he used long division and divided until he got the quotient 0.0638297872, at which poi
babunello [35]
I agree for no reason.
3 0
3 years ago
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