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soldi70 [24.7K]
3 years ago
11

Bacteria can divide every 20 minutes, so 1 bacterium can multiply to 2 in 20 minutes,

Mathematics
1 answer:
8_murik_8 [283]3 years ago
7 0
There will be 2 bacterias after 20 minutes
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Find the coordinates of the midpoint of the segment whose endpoints are H(5, 13) and K(7, 5). (12, 18)
gtnhenbr [62]

\text{The formula of the midpoint of segment AB:}\\\\\left(\dfrac{x_A+x_B}{2};\ \dfrac{y_A+y_B}{2}\right).\\\\\text{We have the points}\ H(5,\ 13)\ \text{and}\ K(7,\ 5).\ \text{Substitute:}\\\\x=\dfrac{5+7}{2}=\dfrac{12}{2}=6\\\\y=\dfrac{13+5}{2}=\dfrac{18}{2}=9\\\\Answer:\ \boxed{(6,\ 9)}

5 0
3 years ago
Opposite reciprocal of -2/3? geometry/algebra
Ulleksa [173]
The answer would be -3/2

A reciprocal is the inverse of a number.
When using this in fractions just switch the numerator and denominator. This is done by dividing the fraction by one, but in the case of a negative fraction you would divide by negative 1.
8 0
3 years ago
Find the smallest solution to the equation 2/3 X² equals 24
iren [92.7K]

Answer:

The smallest solution is -6

Step-by-step explanation:

2/3 x^2 = 24

Multiply each side by 3/2

3/2 *2/3 x^2 = 24*3/2

x^2 = 36

Take the square root of each side

sqrt(x^2) = sqrt(36)

x = ±6

The smallest solution is -6

The largest solution is 6

4 0
3 years ago
Help me asap please
iren [92.7K]

A(-3, -1); B(2, -1), C(2, 3)

You can read the length of AB and BC.

|AB| = 5

|BC| = 4

Use the Pythagorean theorem:

|AC|^2=5^2+4^2\\\\|AC|^2=25+16\\\\|AC|^2=41\to|AC|=\sqrt{41}\\\\|AC|\approx6.4

The perimeter:

P=|AB|+|BC|+|AC|\\\\P=5+4+6.4=15.4

7 0
3 years ago
If α and β are the zeros of the quadratic polynomial f(x) = 3x2–4x + 5, find a polynomialwhose zeros are 2α + 3β and 3α + 2β.
Lelechka [254]

Answer:

\boxed{\sf \ \ \ 3x^2-20x+37\ \ \ }

Step-by-step explanation:

Hello,

a and b are the zeros, we can say that

f(x)=3(x^2-\dfrac{4}{3}x+\dfrac{5}{3}) = 3(x-a)(x-b)=3(x-(a+b)x+ab)

So we can say that

a+b=\dfrac{4}{3}\\ab=\dfrac{5}{3}

Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b

for instance we can write

(x-2a-3b)(x-3a-2b)=x^2-(2a+3b+3a+2b)x+(2a+3b)(3a+2b)\\= x^2-5(a+b)x+6a^2+6b^2+9ab+4ab

and we can notice that

a^2+b^2=(a+b)^2-2ab so

(x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab

it comes

x^2-5*\dfrac{4}{3}x+6(\dfrac{4}{3})^2+\dfrac{5}{3}

multiply by 3

3x^2-20x+2*16+5=3x^2-20x+37

4 0
3 years ago
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