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3 years ago

12

What are the corresponding steps used to find the product of (2x+1)(x+1) ?

1 answer:

expeople1 [14]3 years ago

6 0

You would distribute the 2x to the x and 1= 2x^2+2x
then you would distribute the 1 to the x and 1= 1x+1
then you would add it all together to get
2x^2+3x+1

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A bicyclist goes 43.54 miles in 3.5 hours. What is her speed?

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Divide 43.54 by 3.5
She was going 12.33 miles an hour

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3 years ago

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The life spans, in years, of some plants in a nursery are shown below:

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Jack incorrect plotted the third quartile.

Explanation:

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A colony of bacteria is growing at a rate of 0.2 times its mass. Here time is measured in hours and mass in grams. The mass of t

11Alexandr11 [23.1K]

Answer:

<u>Question 1:</u> $dm/dt=0.2m$ <u />

<u />

<u>Question 2:</u> $m=Ae^{(0.2t)}$ <u />

<u />

<u>Question 3:</u> $m=10e^{(0.2t)}$ <u />

<u />

<u>Question 4:</u> $m=10g$ <u />

Explanation:

<u>Question 1: Write down the differential equation the mass of the bacteria, m, satisfies: m′= .2m</u>

<u></u>

a) By definition: $m'=dm/dt$

b) Given: $rate=0.2m$

c) By substitution: $dm/dt=0.2m$

<u>Question 2: Find the general solution of this equation. Use A as a constant of integration.</u>

a) <u>Separate variables</u>

$dm/m=0.2dt$

b)<u> Integrate</u>

$\int dm/m=\int 0.2dt$

$ln(m)=0.2t+C$

c) <u>Antilogarithm</u>

$m=e^{0.2t+C}$

$m=e^{0.2t}\cdot e^C$

$e^C=A\\\\m=Ae^{(0.2t)}$

<u>Question 3. Which particular solution matches the additional information?</u>

<u></u>

Use the measured rate of 4 grams per hour after 3 hours

$t=3hours,dm/dt=4g/h$

First, find the mass at t = 3 hours

$dm/dt=0.2m\\\\4=0.2m\\\\m=4/0.2\\\\m=20g$

Now substitute in the general solution of the differential equation, to find A:

$m=Ae^{(0.2t)}\\\\20=Ae^{(0.2\times 3)}\\\\A=20/e^{(0.6)}\\\\A=10.976$

Round A to 1 significant figure:

A = 10.

<u>Particular solution:</u>

$m=10e^{(0.2t)}$

<u>Question 4. What was the mass of the bacteria at time =0?</u>

Substitute t = 0 in the equation of the particular solution:

$m=10e^{0}\\\\m=10g$

3 0

3 years ago

There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl

insens350 [35]

We're told that

$P(A\cap B)=0.15$

$P(A\cup B)^C=0.06\implies P(A\cup B)=0.94$

$P(B\mid A)=P(B^C\mid A)=0.5$

where the last fact is due to the law of total probability:

$P(A)=P(A\cap B)+P(A\cap B^C)$

$\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)$

$\implies 1=P(B\mid A)+P(B^C\mid A)$

so that $B\mid A$ and $B^C\mid A$ are complementary.

By definition of conditional probability, we have

$P(B\mid A)=P(B^C\mid A)$

$\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}$

$\implies P(A\cap B)=P(A\cap B^C)$

We make use of the addition rule and complementary probabilities to rewrite this as

$P(A\cap B)=P(A\cap B^C)$

$\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)$

$\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)$

$\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]$

$\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]$

$\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C$

$\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)$

By the law of total probability,

$P(B)=P(A\cap B)+P(A^C\cap B)$

$\implies P(A^C\cap B)=P(B)-P(A\cap B)$

and substituting this into $(*)$ gives

$2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]$

$\implies P(B)=P(A\cup B)-P(A\cap B)$

$\implies P(B)=0.94-0.15=\boxed{0.79}$

8 0

3 years ago

17)Let f(x) = -2x + 5 and g(x) = 9x2 + 4. Find f(8) + g(8) . A)565 B)569 C)564 D)560

siniylev [52]

Answer:

answer B $\boxed{ \ 569 \ }\\$

Step-by-step explanation:

f(8)=-2*8+5=-11

g(8)=9*8*8+4=580

f(8)+g(8)= -11+580=569

4 0

3 years ago