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svetlana [45]
3 years ago
11

2 Complete the statement An B = {...}

Mathematics
1 answer:
slamgirl [31]3 years ago
4 0
B= What’s the question. Maybe try looking at math-way
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Rename the number 780000 = 78
AfilCa [17]
<span>We need to rename the given number which is 780 000 into 78 ________.
=> But before we rename it, we need to find the value of 78
=> Since this is a 6 digit whole number, we have ones, tens, hundreds, thousands, ten thousands, hundred thousands.
=> 7 = hundred thousands
=> 8 = ten thousands
=> 7 hundred thousands + 8 ten thousands
=> Thus, the renamed of 780 000 is 7 hundred 8 ten thousands.

</span>



3 0
3 years ago
Origanal price $60 sale price $45
Blababa [14]

Answer:

di ko po alam ehh

Step-by-step explanation:

sorry po

5 0
3 years ago
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NEED HELP FAST! ALSO 50 POINTS AND BRAINLIEST!! <br><br> What is the value of 5 2/3 (−3.6)?
ratelena [41]

Answer:

-20⅖

Step-by-step explanation:

5⅔(-3.6)

(17/3)(-3.6)

17(-3.6/3)

17(-1.2)

-20.4

-20 ⅖

8 0
2 years ago
What is the value of x?<br><br><br><br> Enter your answer in the box.<br><br> x = ___
loris [4]
(x+4) / 44.8 = 35 / 56
56(x+4) = 35(44.8)
56x + 224 = 1568
56x = 1344
x = 24

answer
x = 24 mm

8 0
3 years ago
When a bactericide is added to a nutrient broth in which bacteria are​ growing, the bacteria population continues to grow for a​
baherus [9]

Answer:

a)  1296 bacteria per hour

b) 0 bacteria per hour

c) -1296 bacteria per hour

Step-by-step explanation:

We are given the following information in the question:

The size of the population at time t​ is given by:

b(t) = 9^6 + 6^4t-6^3t^2

We differentiate the given function.

Thus, the growth rate is given by:

\displaystyle\frac{db(t)}{dt} = \frac{d}{dt}(9^6 + 6^4t-6^3t^2)\\\\= 6^4-2(6^3)t

a) Growth rates at t = 0 hours

\displaystyle\frac{db(t)}{dt} \bigg|_{t=0}= 6^4-2(6^3)(0) = 1296\text{ bacteria per hour}

b) Growth rates at t = 3 hours

\displaystyle\frac{db(t)}{dt} \bigg|_{t=3}= 6^4-2(6^3)(3) = 0\text{ bacteria per hour}

c) Growth rates at t = 6 hours

\displaystyle\frac{db(t)}{dt} \bigg|_{t=6}= 6^4-2(6^3)(6) = -1296\text{ bacteria per hour}

3 0
3 years ago
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