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2 years ago

ben earned 45 dollars mowing lawns. he worked 3.2 hours on monday and 4.3 hours on tuesday. how much did he earn per hour?

Mathematics

1 answer:

dlinn [17]2 years ago

3 0

Answer:

6 dollars per hour

Step-by-step explanation:

We can set up and solve an equation to answer this question. For this problem, let x stand for the amount he earned per hour. The equation that represents this situation is 3.2x + 4.3x = 45. The first term represents monday and the second represents tuesday. When these add up they should equal 45 because that is the total amount he made. To find his pay per hour, solve for x.

First, combine the like terms

7.5x=45

Next, divide both sides by 7.5

x=6

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Antoine Purchase 1.8 kg of apples and 315 dekagrams of oranges This was 1350 g more than the weight of bananas bridges what was

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Answer: 3600 grams

Step-by-step explanation:

A dekagram means 10 grams.

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Therefore the total number of grams when Antoine Purchase 1.8 kg of apples and 315 dekagrams of oranges will be:

= (1.8 × 1000) + (315 × 10)

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= 4950 grams

Since this was 1350g more than the weight of bananas, the weight of the bananas Antonio purchased in grams will be:

= 4950 - 1350

= 3600 grams

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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:

Calculus

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:

Step 1: Define

Identify

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

Step 2: Differentiate

[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

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