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denis-greek [22]
3 years ago
12

Create a sequence in which 1 1/4 is added to each number

Mathematics
1 answer:
RSB [31]3 years ago
5 0
1/4 1/2 3/4 1 1 1/4 1 1/2 1 3/4 2...
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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B3%7D" id="TexFormula1" title="\frac{2}{3}" alt="\frac{2}{3}" align="absmiddle
IceJOKER [234]

\frac{2}{3}  \times  \frac{1}{24}
\frac{1}{3}  \times  \frac{1}{12}
\frac{1}{36}
5 0
4 years ago
PLS HELP ME WITH MATH!!
Nonamiya [84]

Answer:

Jordan's line would go up and Trevor's line would go down. This is because Jordan's like has a positive slope and Trevor's has a negative slope

Step-by-step explanation:

8 0
4 years ago
Read 2 more answers
I’m this geometric sequence, what is the common ratio
Marrrta [24]
ANSWER:

D. - 1 / 2

This is as 104 ÷ - 52 = - 1 / 2

As well as this, - 52 ÷ 26 = - 1 / 2

And this ratio applies for the other numbers in the sequence as well.

Therefore, the answer must be - 1 / 2.

Please mark as brainliest if you found this helpful! :)
Thank you <3
4 0
3 years ago
ASAP Please need help with variables and coefficients, solve all please. First correct answer gets brainliest.
olganol [36]

Variables are the letters in the equation (like x and y)

Coefficients are the numbers before the variables. (like "6" in "6x")

Constants are the actual numbers (like "2" or "1")

Factors: Numbers that are multiplied together (like "5" and "6" in 5 * 6)

5) 4b + 6

Variable: b

Coefficient: 4

Constant: 6

6) 7x + 8h - 3

Variables: x, h

Coefficients: 7, 8

Constants: -3

7) 12 + 2a + 5b

Variables: a, b

Coefficients: 2, 5

Constants: 12

8) 5 + (7 * 8)

Factors: 7 and 8

7 0
3 years ago
The population of fish in a pond from 2001 to 2014 is modeled by the function below where t is the years since 2001. using the f
tresset_1 [31]

Given :

The population of fish in a pond from 2001 to 2014 is modeled by the function below where t is the years since 2001.

p(t) = \dfrac{1125}{1+12e^{-0.17t}}    ...1)

To Find :

Find the number of fish is the pond in 2014.

Solution :

Number of years, t = 2014 - 2001 = 13.

Putting value of t in equation 1), we get :

p(t) = \dfrac{1125}{1+12e^{-0.17\times 13}}\\\\p(t) = 485.67

p(t) ≈ 486

Therefore, number of fish in the pond in 2014 is 486.

Hence, this is the required solution.

7 0
3 years ago
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