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alisha [4.7K]
3 years ago
15

I need it within tomorrow please

Mathematics
1 answer:
Alenkinab [10]3 years ago
6 0

Any even number raised to any power will remain even. (e.g. 2² = 4, 2³ = 8, etc)

Any odd number raised to any power will remain odd. (e.g. 1² = 1³ = ... = 1, 3⁴ = 81, etc)

So (n+1)^{2^k}-n^{2^k} will always be even, since both

[even] - [odd] = [odd]

and

[odd] - [even] = [odd]

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"Sketch a possible graph for a negative Quintic polynomial with one real zero." Can anyone explain how this is done? And if poss
Sidana [21]

quintic means degree of 5.

a negative quintic would be: f(x) = -x⁵ <u>+              </u>

an example with one real zero could be: -x⁵ + x³ + x + 3   (see attachment)

6 0
3 years ago
Read 2 more answers
What is the reciprocal of 4? Enter your answer in the space below, using a slash mark (/) as a fraction bar when needed.​
klio [65]

Answer:

1/4. or 0.25 :))

Step-by-step explanation:

4 as a fraction is 4/1, so the reciprocal is 1/4 :)

4 0
3 years ago
Use a transformation to solve the equation.    f ∙ 8 = 40 a f=5 b f=32 c f=48 df=320
alekssr [168]
Using Taylor expansion, show that
f0
(x0) = f(x0 + h) − f(x0)
h − h
2
f00(ξ),
for some ξ lying in between x0 and x0 + h.
Solution: We expand the function f in a first order Taylor polynomial around x0:
f(x) = f(x0)+(x − x0)f0
(x0)+(x − x0)
2 f00(ξ)
2 ,
where ξ is between x and x0. Let x = x0 + h:
f(x0 + h) = f(x0) + hf0
(x0) + h2
2 f00(ξ).
Solving for f0
(x0), we obtain:
f0
(x0) = f(x0 + h) − f(x0)
h − h
2
f00(ξ)
4 0
3 years ago
Find the solution of the differential equation that satisfies the given initial condition. (dP)/(dt)
kati45 [8]

Answer:

P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2

Step-by-step explanation:

Given

\frac{dP}{dt} = \sqrt{Pt

P(1) = 2

Required

The solution

We have:

\frac{dP}{dt} = \sqrt{Pt

\frac{dP}{dt} = (Pt)^\frac{1}{2}

Split

\frac{dP}{dt} = P^\frac{1}{2} * t^\frac{1}{2}

Divide both sides by P^\frac{1}{2}

\frac{dP}{ P^\frac{1}{2}*dt} = t^\frac{1}{2}

Multiply both sides by dt

\frac{dP}{ P^\frac{1}{2}} = t^\frac{1}{2} \cdot dt

Integrate

\int \frac{dP}{ P^\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt

Rewrite as:

\int dP \cdot P^\frac{-1}{2} = \int t^\frac{1}{2} \cdot dt

Integrate the left hand side

\frac{P^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} = \int t^\frac{1}{2} \cdot dt

\frac{P^{\frac{-1}{2}+1}}{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt

2P^{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt

Integrate the right hand side

2P^{\frac{1}{2}} = \frac{t^{\frac{1}{2} +1 }}{\frac{1}{2} +1 } + c

2P^{\frac{1}{2}} = \frac{t^{\frac{3}{2}}}{\frac{3}{2} } + c

2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c ---- (1)

To solve for c, we first make c the subject

c = 2P^{\frac{1}{2}} - \frac{2}{3}t^\frac{3}{2}

P(1) = 2 means

t = 1; P =2

So:

c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1^\frac{3}{2}

c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1

c = 2\sqrt 2 - \frac{2}{3}

So, we have:

2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c

2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + 2\sqrt 2 - \frac{2}{3}

Divide through by 2

P^{\frac{1}{2}} = \frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3}

Square both sides

P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2

6 0
3 years ago
A = 1/3, b = 9, c = 5, and d = 10.<br><br> 12a + c - b
emmasim [6.3K]

Answer:

The equation equals <em>0</em>

Step-by-step explanation:

<u>12 ( 1/3 )</u> + 5 - 9

<u>4 + 5</u> - 9

<u>9 - 9</u>

= 0

5 0
3 years ago
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