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2 years ago

write a number from 20 to 50 that has both tens and ones. use pictures and words to show the tens and ones.

Mathematics

2 answers:

Vikki [24]2 years ago

6 0

Answer:

Step-by-step explanation:

This problem is asking for a number between 20 and 50 that has tens and ones, it could be 25, 35 or 45. Basically, the answer could be all numbers that doesn't end in zero or one, because zero means that there aren't any ones, and if ends in one, that means it only has one one.

So, the answer we choose is 45, which has 4 tens and 5 ones, that is, 4 tens that equals 40, and 5 ones that equals 5.

slavikrds [6]2 years ago

5 0

The answer is 35 because it in beetween 20 and 50

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Help differentiate this

Arte-miy333 [17]

Answer:

$\displaystyle y' = 20x^3 + 6x^2 + 70x + 9$

General Formulas and Concepts:

Pre-Algebra

Distributive Property

Algebra I

Terms/Coefficients
Expand by FOIL
Functions
Function Notation

Calculus

Derivatives

Derivative Notation

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

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

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

$\displaystyle y = (x^3 + 7x - 1)(5x + 2)$

Step 2: Differentiate

Product Rule: $\displaystyle y' = \frac{d}{dx}[(x^3 + 7x - 1)](5x + 2) + (x^3 + 7x - 1)\frac{d}{dx}[(5x + 2)]$
Basic Power Rule [Derivative Property - Addition/Subtraction]: $\displaystyle y' = (3x^{3 - 1}+ 7x^{1 - 1} - 0)(5x + 2) + (x^3 + 7x - 1)(5x^{1 - 1} + 0)$
Simplify: $\displaystyle y' = (3x^2+ 7)(5x + 2) + 5(x^3 + 7x - 1)$
Expand: $\displaystyle y' = 15x^3 + 6x^2 + 35x + 14 + 5(x^3 + 7x - 1)$
[Distributive Property] Distribute 5: $\displaystyle y' = 15x^3 + 6x^2 + 35x + 14 + 5x^3 + 35x - 5$
Combine like terms: $\displaystyle y' = 20x^3 + 6x^2 + 70x + 9$