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Leni [432]
3 years ago
15

AABC ~ DEF. What sequence of transformations will move AABC onto ADEF

Mathematics
1 answer:
Taya2010 [7]3 years ago
8 0

Answer:

C

Step-by-step explanation:

i am smart

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The South Coast Air Basin (Los Angeles, Orange, and San Bernardino counties) does not meet the air quality standards for carbon
Archy [21]

Answer:

0.2225 gr/mile

Step-by-step explanation:

Let's work out first the amount of CO sent out in 1990.

The population we estimated in 12.5 million people with a total of driven miles per year of about  

12.5 million*8,700 = 108,750 million miles.

With an CO emission factor of 0.9 g per mile, we would have a total of CO emitted rounding 0.9*180,750 = 97,875 million grams

Now, we must estimate the population for 2020.

Since we are assuming an exponential growth, the population in year t is given by a function

\bf P(t)= Ce^{kt}

where C and k are constants to be determined.

We can take 1980 as year 0. This way calculations are lighter. 1990 is year 10 and 2020 is year 20.

So P(0) = 10.3 and C=10.3

So far we have

\bf P(t)= 10.3e^{kt}

Given that P(10)=12.5

\bf 10.3e^{k*10}=12.5\rightarrow e^{10k}=\frac{12.5}{10.3}=1.21359\rightarrow \\10k=log(1.21359)\rightarrow k=0.01936

And the function that models the population growth is

\bf P(t)= 10.3e^{0.01936t}

We need P(20)

\bf P(20)= 10.3e^{0.01936*20}=10.3e^{0.38717}=15.17\;million

If the miles driven per person per year remains constant at 8700 mi/yr.person, then we have a total miles driven of

15.17*8,700=131,979 million miles, so the CO emitted would be 0.9*131,979=118,781.1 million grams.

The 30% of the CO sent out in 1990 is 0.3*97,875=29,362.5 million grams.

We must reduce 118,781.1 down to 29,362.5

Hence the new CO emission factor would be

29,362.5/131,979 = 0.2225 gr/mile

6 0
3 years ago
Describe the transformation. 
IRISSAK [1]
C I guess I was learning this in class but I'm not sure.
5 0
3 years ago
Read 2 more answers
a car drives 50 kilometres east and then 40 kilometres due north. how far is the car from its starting point?
ozzi

Answer:

90 kilometres

Step-by-step explanation:

6 0
2 years ago
Solve the following quadratic equation 6x^2-5x-4=0
jeka94

Answer:

=

−

±

2

−

4

√

2

x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}

x=2a−b±b2−4ac​​

Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.

6

2

−

5

−

4

=

0

6x^{2}-5x-4=0

6x2−5x−4=0

=

6

a={\color{#c92786}{6}}

a=6

=

−

5

b={\color{#e8710a}{-5}}

b=−5

=

−

4

c={\color{#129eaf}{-4}}

c=−4

=

−

(

−

5

)

±

(

−

5

)

2

−

4

⋅

6

(

−

4

)

√

2

⋅

6

Step-by-step explanation:

6 0
2 years ago
Math help 30 points
mash [69]

Again here we're using like terms

a) -2x² + 12x² = 10x² (they have the same term that's why I grouped them together)

-4x + 2x = -2x

13 - 25 = -12

So the final answer is 10x² - 2x -12

b) 7x² - (-7x²) = 7x² + 7x² or 14x²

4x - (-3x) = 4x + 3x or 7x

-26 - 15 = -41

The final answer to this one is 14x² + 7x - 41

7 0
2 years ago
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