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3 years ago

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Help please! Explain in detail how to get a microscope into focus, working your way up to using the high power lens. Include any

safety Issues you need to be aware of.

1 answer:

Amiraneli [1.4K]3 years ago

5 0

Well
You take the thing

You make sure a very good amount of light is pointing at the microscope

And then you turn the zoom-in or zoom-out

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A ball, thrown vertically upwards, from the ground, has its height h (in meters) expressed as a function of time t (in seconds),

Ivenika [448]

Answer:

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Step-by-step explanation:

fddefefgerf

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3 years ago

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3 and 1/3 + 4 and 5/7 =

klasskru [66]

The answer is 8 and 1/21 in fixed fraction.

and decimal form it would be 8.047619

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2 years ago

At the market, 9 apples cost $4.00. How much does 10 apples cost?

ycow [4]

Answer:

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Step-by-step explanation:

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3 years ago

If f(x) = 9x10 tan−1x, find f '(x).

djverab [1.8K]

Answer:

$\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}$

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle f(x) = 9x^{10} \tan^{-1}(x)$

<u>Step 2: Differentiate</u>

[Function] Derivative Rule [Product Rule]: $\displaystyle f'(x) = \frac{d}{dx}[9x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle f'(x) = 9 \frac{d}{dx}[x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Basic Power Rule: $\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Arctrig Derivative: $\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

7 0

3 years ago

Points F and E on the coordinate grid below show the positions of two midfield players of a soccer team:

Lelechka [254]

Answer:

F is E reflected across the y-axis; only the signs of the x-coordinates of F and E are different.

Step-by-step explanation:

The question is incomplete. Here is the complete question.

" Points F and E on the coordinate grid below show the positions of two midfield players of a soccer team:

Coordinate grid shown from negative 4 to positive 4 on x-axis and negative 4 to positive 4 on y-axis. From the origin, point E is 1.5 units to the left and 3 units up. From the origin, point F is 1.5 units to the right and 3 units up.

Which statement best describes the relationship between the positions of the two midfield players? (1 point)

Based on the coordinate of E;

The value on the x axis is -1.5 since the point is 1.5unitsTO THE LEFT from the origin(-ve x-axis acts towards the left in space). The value on the y-axis will be +3units since its 3units UP(+ve y axis acts upward).

The E coordinates will therefore be (-1.5, 3)

Similarly, for coordinate F:

If from the origin, point F is 1.5 units TO THE RIGHT, then the x-axis will be +1.5 (x-axis is positive towards the right) and +3 units up(+ve y axis acts upward).

The F coordinate will therefore be;

(1.5, 3).

From both coordinates, it can be seen that both x axis are of the same value but different signs. This means that the points E and F will act in the opposite side of the y-axis and a reflector of each other.

Based on the explanation above, it can be concluded that "F is E reflected across the y-axis; only the signs of the x-coordinates of F and E are different."

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3 years ago