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kobusy [5.1K]
3 years ago
7

the second term in a geometric sequence is 20. the fourth term in the same sequence is 45/4 or 11.25. what is the common ratio

Mathematics
1 answer:
slega [8]3 years ago
5 0
4th term / second term  = a1r^3 / a1r = r^2

so r^2  = 45/4 / 20 = 45/80  = 9/16

so r = sqrt 9/16  = 3/4  answer
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In the expression 852.763 x 10?, what is the missing exponent that would result in a number that is 1,000 times the value of 852
Molodets [167]

Answer:

The missing exponent = power of 3

852.763 x 10³ = 852763

Step-by-step explanation:

The expression 852.763 x 10?

A number that is 1,000 times the value of 852.763 is calculated as:

852.763 × 1000 = 852763

Converting this to exponent

852.763 × 10³ = 852763

Therefore, the missing exponent that would result in number that is 1,000 times the value of 852.763 is power of 3 meaning 10³

7 0
3 years ago
Can someone help me pls pls help with this question
True [87]

Answer:

D.

Step-by-step explanation:

The mean is 11,000

8 0
2 years ago
Can someone plz help me ty
Ostrovityanka [42]

Answer:

8/15

Step-by-step explanation:

you first want to change your whole number, 4, into a fraction so all you have to do is add a 1 in the denominator so it will be 4/1

then you multiply across

2/15 * 4/1

you first multiply 2 and 4 and your answer will be 8 which is the numerator

then multiply 15 and 1 which is 15, the denominator

then you get

8/15

5 0
3 years ago
dillon pays x dollars to get a haircut. he gives the stylist a 15% tip. the expression representing his total cost is x + 0.15x
bagirrra123 [75]

Answer:

B) 1.15x, adding 15% to the cost of haircut is same as to multiply cost by 1.15

Step-by-step explanation:

Given expression = x + 15% of x

x + (15/100) x = x + 0.15x

= (1+ 0.15) x = 1.15x

15% = 0.15 , added to x = 1.15x

4 0
3 years ago
Please help me with this! I will mark you as brainliest!
erastovalidia [21]

Step-by-step explanation:

\frac{ {5}^{3} }{ {5}^{7} }

\frac{1}{ {5}^{7 - 3} }

\frac{1}{ {5}^{4} }

or

{5}^{ - 4}

Hope it will help :)

4 0
3 years ago
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