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3 years ago

the stove and the freezer weigh at least 260kg. the stove weighs 115 kg. How much would the freezer weigh?

Mathematics

2 answers:

Aleks [24]3 years ago

7 0

About 145 I gonna guess I dont know if that's right

saveliy_v [14]3 years ago

7 0

Answer:

freezer weight = at least 145 kg

Step-by-step explanation:

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Give an example of an absolute value inequality whose answers falls between two values. Give an example of an absolute value ine

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Answer:

A) Suppose we have an onedimensional situation.

in the 0 of our x-axis, we have a fruit tree, and we want to rest at a distance no bigger than 6 ft of the tree, then all the possible positions of our resting place are:

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we can write this as: IxI < 6ft

b) now we think the opposite situation, we want to rest at least 6ft away from the tree, then we have that:

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3 years ago

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Anuta_ua [19.1K]

Answer:

Step-by-step explanation:

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I guess I'm lacking in differential equations. I couldn't solve this question. Can you help me?

Sonja [21]

Answer:

See Explanation.

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Reciprocals

Algebra II

Log/Ln Property: $ln(\frac{a}{b} ) = ln(a) - ln(b)$

Calculus

Derivatives

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Derivative of Ln: $\frac{d}{dx} [ln(u)] = \frac{u'}{u}$

Step-by-step explanation:

Step 1: Define

$ln(\frac{2x-1}{x-1} )=t$

Step 2: Differentiate

Rewrite: $t = ln(\frac{2x-1}{x-1})$
Rewrite [Ln Properties]: $t = ln(2x-1) - ln(x - 1)$
Differentiate [Ln/Chain Rule/Basic Power Rule]: $\frac{dt}{dx} = \frac{1}{2x-1} \cdot 2 - \frac{1}{x-1} \cdot 1$
Simplify: $\frac{dt}{dx} = \frac{2}{2x-1} - \frac{1}{x-1}$
Rewrite: $\frac{dt}{dx} = \frac{2(x-1)}{(2x-1)(x-1)} - \frac{2x-1}{(2x-1)(x-1)}$
Combine: $\frac{dt}{dx} = \frac{-1}{(2x-1)(x-1)}$
Reciprocate: $\frac{dx}{dt} = -(2x-1)(x-1)$
Distribute: $\frac{dx}{dt} = (1-2x)(x-1)$