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defon
3 years ago
12

Bacteria vary somewhat in size, but a diameter of 1.5 μ m is not unusual.

Mathematics
1 answer:
Dmitrij [34]3 years ago
6 0

Answer:

a) 1.8 × 10^-12 cm³ or 1.8 × 10^-12 cubic meters

b) 7.1 × 10^-6 mm² or 7.1 × 10^-6 square millimeters

Step-by-step explanation:

a) We are assuming that the shape of the bacteria is a sphere.

Hence, Volume of the Sphere(Bacteria) = 4/3 × π × r³

Diameter = 1.5 μm

Radius = Diameter/2 = 1.5μm/2

= 0.75μm

We are told that the volume should be in cubic centimeters

Converting 0.75μm to centimeters

1 μm = 1 × 10^-4 cm

0.75 μm =

Cross Multiply

= 0.75 μm × 1 × 10^-4 cm/ 1 μm

= 0.000075cm

Volume of the Sphere(Bacteria) = 4/3 × π × r³

= 4/3 × π × (0.000075)³

= 1.767145867 × 10^-12 cm³

Approximately as 2 significant figures = 1.8 × 10^-12 cm³

b) The formula for the Surface area of a Sphere = 4πr²

Diameter = 1.5 μm

Radius = Diameter/2 = 1.5μm/2

= 0.75μm

We are told that the surface area should be in square millimeters

Converting 0.75μm to millimeters

1 μm = 0.001 mm

0.75 μm =

Cross Multiply

= 0.75 μm ×0.001mm/ 1 μm

= 0.00075mm

Surface Area of a Sphere

= 4 × π × r²

= 4 × π × 0.00075²

= 7.06858 ×10^-6 mm²

Approximately to 2 significant figures

= 7.1 × 10^-6 mm²

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If you simplify each of the numbers in A, B, and C, you will find that they will come to equal 7:5

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A pair of jeans is discounted by 12% and then a further 17% discount is
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Answer:

$171.14

Step-by-step explanation:

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Is the following a function ?
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A model of a famous statue is 2 1/2 inches tall. The actual statue is 3 1/3 feet tall. What is the ratio of the height of the mo
Gekata [30.6K]

1. A model of a famous statue is 2\dfrac{1}{2} inches tall that is

\dfrac{2\cdot 2+1}{2}=\dfrac{5}{2} in.

2. The actual statue is 3\dfrac{1}{3} feet tall that is

\dfrac{3\cdot 3+1}{3}=\dfrac{10}{3}\ ft=\dfrac{10}{3}\cdot 12=40\ in.

3. The ratio of the height of the model to the height of the actual statue in simplest form is

\dfrac{\dfrac{5}{2}}{40}=\dfrac{5}{2}\cdot \dfrac{1}{40}=\dfrac{1}{16}.

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Rewrite the system of linear equations as a matrix equation AX = B.
iren2701 [21]

Answer:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]

And the vector of vriables (X) is:

\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

In conclusion, AX=B is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

7 0
3 years ago
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