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2 years ago

Which comparison is correct? |5| > 7 7 < |7| |-3| > |-4| -5 < -4

Mathematics

1 answer:

Bas_tet [7]2 years ago

5 0

Answer:

-5 < - 4 is correct

Step-by-step explanation:

I5I > 7 is wrong because I5I < 7

7 < I7I is wrong . 7 = I7I

I-3I > I-4I is wrong because absolute value is always positive.

So I-3I < I-4I

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what is the equation for “the product of a number and 3 is 5 less than the quotient of a number and 4

marishachu [46]

Equation 1.
20÷4=5
0×3=0
5-5=0
Equation 2.
44÷4=11
2×3=6
11-5=6
Equation 3.
56÷4=14
3×3=9
14-5=9

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3 years ago

Mr. Wong spent 2/3 of his workday selling products and the rest of his workday making phone calls. What fraction of his workday

tensa zangetsu [6.8K]

Answer:

1/3

Step-by-step explanation:

if he has spent 2/3 of his workday selling products that is almost his whole day. He has 1/3 of his workday left so he dedicated it to making calls for his company.

So you work if you subtract 2/3 from a whole that leaves you with 1/3 of your day left

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3 years ago

Read 2 more answers

5/1/10x - 6/4/5 = 3/1/4

emmasim [6.3K]

I think the answer is A

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3 years ago

Read 2 more answers

The product of twice a and b

Harman [31]

The product to your question is 2a+b

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3 years ago

Find the derivative.

krek1111 [17]

Answer:

$\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

General Formulas and Concepts:

Algebra I

Terms/Coefficients

Expanding/Factoring

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = \frac{\sqrt{x}}{e^x}$

Step 2: Differentiate

Derivative Rule [Quotient Rule]: $\displaystyle f'(x) = \frac{(\sqrt{x})'e^x - \sqrt{x}(e^x)'}{(e^x)^2}$
Basic Power Rule: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}(e^x)'}{(e^x)^2}$
Exponential Differentiation: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{(e^x)^2}$
Simplify: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{e^{2x}}$
Rewrite: $\displaystyle f'(x) = \bigg( \frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x \bigg) e^{-2x}$
Factor: $\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

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

Unit: Differentiation

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2 years ago