I forgot to study sorry guys

Mathematics

1 answer:

goldenfox [79]2 years ago

5 0

I think that the answer is A) because the range is y.

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A library is open til' 5:00 P.M. On Wednesdays, the library closes 2 hours early. How many hours for a month is the library open

svet-max [94.6K]

So
if you assume that the month has 30 days and that the library opens at midnight, then
24 hours in a day
5 pm=12+5=17 hours
on wednessday=17-2=15 hours
wednessday=1/7 of week
so we find 1/7 of 30 which is 30/7=4 and 2/7
then subtract that from 30
30-4 and 2/7=25 and 5/7

ok so then we have
25 and 5/7 days is 17 hours and
4 and 2/7 days is 15 hours
so just multipy them and add

25 and 5/7 times 17=437.143 hours
4 and 2/7 days times 15 =64.2857
add

437.143+64.2856=501.429
so aprox 501.429

the real equation is $[(\frac{1}{7})(n)(15)]+[ (\frac{6}{7}) (n)(17)]=hours$ where n represents the number of days in the month

apros 501.429

8 0

2 years ago

20000×50×45×0×23×5×6+229+466+7+5+8+9+1+2+3+45+6+7+8+9+10. Hope you like question don't answer if you don't want to. Have a great

Lemur [1.5K]

815? but have a great day as well!

4 0

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