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m_a_m_a [10]
2 years ago
14

What equation does this represent?

Mathematics
1 answer:
Fynjy0 [20]2 years ago
5 0
Linear… do not take my word for that
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Which statement best summarizes this passage?
ololo11 [35]
Say; Alexander the Great continued exploring, despite his Greek soldiers’ quitting
7 0
4 years ago
3. Explain why the polynomial 34 + 5x3 - 7x2 + 3x has a degree 3 and not 4.​
Pepsi [2]
2x6 + 4x5 + x4 + 11x3 + 2x2 + 4x + 4
7 0
3 years ago
Read 2 more answers
The track team is trying to reduce their time for a relay race. First they reduce their time by 1.2 minutes. Then they are able
Digiron [165]

Answer:

5.2 Minutes, or 5 minutes and 12 seconds

Step-by-step explanation:

You can solve this problem by working backward.  Since our number gets smaller by a 1/10 or 0.1, in order to work backward we must multiply 3.60 by 10/9, since one tenth has been taken away already.  When we do this, we get 4.0.  When we add 1.2 minutes to that, we get a final answer of 5.2 minutes, or 5 minutes and 12 seconds.

5 0
3 years ago
Probability question, which I am stuck on!
son4ous [18]

Answer:

land on 3: 36 times

land on 4: 63 times

Step-by-step explanation:

A biased dice is the opposite of a fair dice.

A fair dice has the same probability of landing any of the six numbers: 1/6

The biased dice has different probabilities for its results.

To solve this question, first we need to find the probability of landing a 3.

The sum of all probabilities need to be 1, so:

0.13 + 0.05 + p(3) + 0.21 + 0.19 + 0.3 = 1

p(3) = 1 - 0.88 = 0.12

If we roll the dice 300 times, the expected number of times the dice will land:

on 3: 300 * p(3) = 300 * 0.12 = 36 times

on 4: 300 * p(4) = 300 * 0.21 = 63 times

7 0
3 years ago
Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0
Leokris [45]

Substitute v(x)=-2x+y(x), so that \dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}. Then the ODE is equivalent to

\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7

which is separable as

\dfrac{\mathrm dv}{v^2-9}=\mathrm dx

Split the left side into partial fractions,

\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)

so that integrating both sides is trivial and we get

\dfrac{\ln|v-3|-\ln|v+3|}6=x+C

\ln\left|\dfrac{v-3}{v+3}\right|=6x+C

\dfrac{v-3}{v+3}=Ce^{6x}

\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}

\dfrac6{v+3}=1-Ce^{6x}

v=\dfrac6{1-Ce^{6x}}-3

-2x+y=\dfrac6{1-Ce^{6x}}-3

y=2x+\dfrac6{1-Ce^{6x}}-3

Given the initial condition y(0)=0, we find

0=\dfrac6{1-C}-3\implies C=-1

so that the ODE has the particular solution,

\boxed{y=2x+\dfrac6{1+e^{6x}}-3}

5 0
3 years ago
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