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3 years ago

68 more than the quotient of an unknown number and 54 is 72.

2 answers:

4 0

68 + (x / 54) = 72 68 + (x / 54) • 54 = 72 • 54 68 + x = 72 • 54 68 + x = 3,888 68 + x - 68 = 3,888 - 68 x = 3,888 - 68 x = 3,820 I'm pretty sure this is right. Good luck!

rosijanka [135]3 years ago

4 0

The unknown number is 216

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Question 5 of 10

maxonik [38]

Step-by-step explanation:

(f + g)(x) = f(x) + g(x) = (4x² + 1) + (x² - 5) = 5x² - 4. (D)

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3 years ago

Read 2 more answers

A rumor spreads through a small town. Let y ( t ) be the fraction of the population that has heard the rumor at time t and assum

Ivan

Answer:

Differential equation

$\frac{dy}{dt} =ky(1-y)$

Solution

$y=\frac{1}{1+4e^{-0.327t}}$

Value of constant k=0.327 days^(-1)

The rumor reaches 80% at 8.48 days.

Step-by-step explanation:

We know

y(t): proportion of people that heard the rumor

y'(t)=ky(1-y), rate of spread of the rumor

Differential equation

$\frac{dy}{dt} =ky(1-y)$

Solving the differential equation

$\frac{dy}{y(1-y)}=k\cdot dt \\\\\int \frac{dx}{y(1-y)} =k \int dt \\\\-ln(1-\frac{1}{y} )+C_0=kt\\\\1-\frac{1}{y} =Ce^{-kt}\\\\\frac{1}{y} =1-Ce^{-kt}\\\\y=\frac{1}{1-Ce^{-kt}}$

Initial conditions:

$y(0)=0.2\\y(3)=0.4\\\\y(0)=0.2=\frac{1}{1-Ce^0}\\\\1-C=1/0.2\\\\C=1-1/0.2= -4\\\\\\y(3)=0.4=\frac{1}{1+4e^{-3k}} \\\\1+4e^{-3k}=1/0.4\\\\e^{-3k}=(2.5-1)/4=0.375\\\\k=ln(0.375)/(-3)=0.327\\\\\\y=\frac{1}{1+4e^{-0.327t}}$

Value of constant k=0.327 days^(-1)

At what time the rumor reaches 80%?

$y(t)=0.8=\frac{1}{1+4e^{-0.327t}} \\\\1+4e^{-0.327t}=1/0.8=1.25\\\\e^{-0.327t}=(1.25-1)/4=0.0625\\\\t=ln(0.0625)/(-0.327)=8.48$

The rumor reaches 80% at 8.48 days.

8 0

3 years ago

You have been cleanin diesel dispensers for 35 minutes you have completed 6 you hace completed 6 how many minutes its taken to c

ololo11 [35]

The answer is 5.833333333 or 5.83 minutes

7 0

3 years ago

How many distinct pairs of disjoint non-empty subsets of A are there, the union of which is all of A?

Mrac [35]

A = {1, 2, 5, 6, 8}
{1} U {2, 5, 6, 8}
{2} U {1, 5, 6, 8}
{5} U {1, 2, 6, 8}
{6} U {1, 2, 5, 8}
{8} U {1, 2, 5, 6}
{1, 2} U {5, 6, 8}
{1, 5} U {2, 6, 8}
{1, 6} U {2, 5, 8}
{1, 8} U {2, 5, 6}
{1, 2, 5} U {6, 8}
{1, 2, 6} U {5, 8}
{1, 2, 8} U {5, 6}
{1, 5, 6} U {2, 8}
{1, 5, 8} U {2, 6}
{1, 6, 8} U {2, 5}
The answer is 15 distinct pairs of disjoint non-empty subsets.

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3 years ago

Which set of data has the strongest linear association?

Inga [223]

The first one because it goes through pretty much all the points on the graph.

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3 years ago