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defon
3 years ago
11

Use repeated subtraction to divide.1÷1/5

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
7 0

1 - 1/5 = 5/5 - 1/5 = 4/5

4/5 - 1/5 = 3/5

3/5 - 1/5 = 2/5

2/5 - 1/5 = 1/5

1/5 - 1/5 = 0

We can subtract 1/5 from 1 a total of 5 times, so 1 ÷ 1/5 = 5.

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1 2 3 4 5 6 7 8 9 10
Sunny_sXe [5.5K]

Answer:

x^{2} + 2

Step-by-step explanation:

Group:

(2x^{3} + 4x) (-5x^{2} - 10)

2x (x^{2} + 2)     -5 (x^{2} + 2)

Common binomial factor = (x^{2} + 2)

4 0
3 years ago
Read 2 more answers
Solve for x 2^x+6=2^5
solong [7]

Answer:

Stesksp-by-step explanation:

6 0
3 years ago
lynne took a taxicab from her office to the airport. she had to pay a flat fee of $2.05 plus $0.90 per mile. the total cost was
notsponge [240]
2.05 + 0.90m = 5.65
0.90m = 5.65 - 2.05
0.90m = 3.60
m = 3.60 / 0.90
m = 4 <=== the taxi trip was 4 miles
5 0
3 years ago
Read 2 more answers
I NEED HELP ASAP !!!!!!!!!!!!!!!!!!!!
Anon25 [30]

1) A: The first table

2) B: The water level increases by 2 inches each hour.

Hope this helps

8 0
3 years ago
Read 2 more answers
On a particular production line, the likelihood that a light bulb is defective is 10%. seven light bulbs are randomly selected.
amid [387]

Answer:

0.9995

Step-by-step explanation:

10% = 0.10

1 - 0.10 = 0.9

n = number of light bulbs = 7

we calculate this using binomial distribution.

p(x) = nCx × p^x(1-p)^n-x

our question says at most 4 is defective

= (7C0 × 0.1⁰ × 0.9⁷) + (7C1 × 0.1¹ × 0.9⁶) + (7C2 × 0.1² × 0.9⁵) + (7C3 × 0.1³ × 0.9⁴) + (7C4 × 0.1⁴ × 0.9³)

= 0.478 + 0.372 + 0.1239 + 0.023 + 0.0026

= 0.9995

we have 0.9995 probability that at most 4 light bulbs are defective.

6 0
3 years ago
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