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sergij07 [2.7K]
3 years ago
8

there are 16 bactiria in a beaker and four hours later there are 228 bacteria in the beaker. what is the rate of change per hour

in the number of bactiria
Mathematics
2 answers:
Dominik [7]3 years ago
8 0

Write and solve the appropriate exponential equation:

228 = 16(r)^4, where r is the change per hour in the number of bacteria:

228

------ = r^4     =   14.25.  

 16

Taking the fourth root of both sides, we get:  r = 1.94 (answer)

Sav [38]3 years ago
8 0

Initial number of bacteria = 16

Final number of bacteria after 4 hours = 228

So, bacteria increased by = 228-16=212

So per hour the bacteria increased by = \frac{212}{4} = 53

Hence, the bacteria increased by 53 per hour.

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X^1/4 +3 =0 -Solve the equation. Check your solutions
mina [271]

Answer:

81

Step-by-step explanation:

subtract 3 from both sides->x^1/4=-3

set both sides to the power of 4-> x=(-3)^4

solve-> x=81

3 0
3 years ago
Which statement is true?
Veronika [31]

Answer:

B. Only 140 is an outlier

Step-by-step explanation:

To properly identify an outlier, you must first know what it is. An outlier is a number that is either a lot higher or a lot lower than the average in a set of numbers. For example, if you had a number set of 1, 3, 4, 6, and 72, you can deduce that 72 is the outlier because it's very far away compared to the other numbers in the set.

In the set that's provided, the numbers tend to range in the double digits, going up in small increments from 15 to 89. However, we can see that 140 is a lot higher than the rest of the numbers in the set, so we can assume that 140 is an outlier.

7 0
3 years ago
Read 2 more answers
( 16.9 - 5.47 ) x 7.09 plzz i want with full solution
Mila [183]

Answer:

81.03

Step-by-step explanation:

(16.9-5.47)×7.09

(11.43)×7.09

81.0387

OR

81.03

HOPE THIS HELPS YOU

8 0
3 years ago
30 POINTS
vivado [14]
Centre= (-1,-2)
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All working shown on photo

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B
riadik2000 [5.3K]

Let

S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n

where we assume |r| < 1. Multiplying on both sides by r gives

r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}

and subtracting this from S_n gives

(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}

As n → ∞, the exponential term will converge to 0, and the partial sums S_n will converge to

\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}

Now, we're given

a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a

a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}

We must have |r| < 1 since both sums converge, so

\dfrac{15}a = \dfrac1{1-r}

\dfrac{150}{a^2} = \dfrac1{1-r^2}

Solving for r by substitution, we have

\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)

\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}

Recalling the difference of squares identity, we have

\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}

We've already confirmed r ≠ 1, so we can simplify this to

\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15

It follows that

\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12

and so the sum we want is

ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0
2 years ago
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