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2 years ago

Subtract −7n+2 from 2n−1 .

Mathematics

2 answers:

lakkis [162]2 years ago

5 0

2n−1 + 7n - 2
= 9n - 3

-------------------

Aliun [14]2 years ago

5 0

2n - 1 - (-7n + 2) =
2n - 1 + 7n - 2 =
9n - 3 <===

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0.1+0.2+0.3×0.4????

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2 years ago

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If f(x) = 3x^2 - 4x + 5 and g(x) = 2x^2 + 2, find f(x) – g(x). Answer: x^2 - 4x + 3

Nadusha1986 [10]

Answer:

f(x) – g(x) = x^2 - 4x + 3

Step-by-step explanation:

f(x) = 3x^2 - 4x + 5 and g(x) = 2x^2 + 2,

f(x) – g(x) = 3x^2 - 4x + 5 - ( 2x^2 + 2)

Distribute the minus sign

f(x) – g(x) = 3x^2 - 4x + 5 - 2x^2 - 2

Combine like terms

f(x) – g(x) = x^2 - 4x + 3

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2 years ago

| 2x + 2y = 10 y = 5+x

melisa1 [442]

2x + 2(5+x)=10
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2 years ago

A. 6 b 1/6 c 4/16 d 1/4

Dmitrij [34]

D. 1/4
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2 years ago

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

Subtract −7n+2 ​​from 2n−1 ​.

Subtract −7n+2 from 2n−1 .