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Usimov [2.4K]
3 years ago
5

Help is appreciated! What is the slope of the line, and what is the y-intercept?

Mathematics
2 answers:
blsea [12.9K]3 years ago
8 0
Slope is 1 and the y intercept is (0,-7)
Lady_Fox [76]3 years ago
3 0
The slope is 1
The y-int is (0,-7)
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What's the difference between 5.4 x 10-7 and 7.1 x 10-8? Express your answer
vichka [17]

Answer:

.76 x 10 or 7.6 x 10^0

Step-by-step explanation:

divide coefficients and subtract exponents

5.4 / 7.1 = .76

10^-7 / 10^-8 = 10

5 0
3 years ago
A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 40° F. Af
fgiga [73]

Answer:

53.3324

Step-by-step explanation:

given that a  thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 40° F.

By Newton law of cooling we have

T(t) = T+(T_0-T)e^{-kt}

where T (t) is temperature at time t,T =surrounding temperature = 40, T0 =70 = initial temperature

After half minute thermometer reads 60° F. Using this we can find k

T(0,5) = 40+(70-40)e^{-k/2} = 60\\e^{-k/2} =2/3\\-k/2 = -0.4055\\k = 0.8110

So equation is

T(t) = 40+(30)e^{-0.8110t}\\

When t=1,

we get

T(1) = 40+(30)e^{-0.8110}\\\\=53.3324

5 0
3 years ago
Imagina que además de su trabajo en la bi biblioteca ,lizzie planea trabajar como voluntaria 3 horas por semana en un refugio de
Tamiku [17]

Answer:

(20 + 5w) + (10 + 3( w - 1 )) = 27 +8w

Step-by-step explanation:

¡espero que esto ayude!

3 0
3 years ago
Question B,C and D I need answers to these parts.
Alik [6]
Explanation<h2>Part b</h2>

So in part (b) we must give the annual growth factor. For any quantity that increases annualy, this factor is the number by which the quantity multiplies itself every year. For a given quantity that increases at a rate of p% its growth factor is given by:

1+\frac{p}{100}

In this case the production of ethanol increases at a rate of 5.5% per year which means that the growth factor is:

1+\frac{5.5}{100}=1.055

Then the growth factor and answer to part (b) is 1.055.

In part (c) we must write an equation that models the increase in the ethanol production. Given an initial qunatity Q and an annual growth factor k the quantity after x years is given by multiplying the initial quantity by the growth factor raised to the x. Then we get:

C(x)=Q\cdot k^x

In this case we need a function for the amount of ethanol produced E, x years after 1990. The 0.9 billion gallons produced that year compose our initial quantity and knowing that the annual growth factor is 1.055 we have the desire equation:

E(x)=0.9\cdot1.055^x

And that's the answer to part (c).

In part (d) we have to compare the actual production of ethanol in 2008 to the one predicted by the equation of part (c). Remember that x represents the years passed after 1990 so for the year 2008 the value of x is given by:

x=2008-1990=18

So the number we are looking for is E(18):

E(18)=0.9\cdot1.055^{18}=2.359

So the production estimated for 2008 is 2.359 billions of gallons. Then the actual production of 9 billion gallons is higher than the one predicted by our equation.

5 0
1 year ago
ANSWER QUICK PLS ANSWER QUICK PLS
il63 [147K]

Answer:

3

Step-by-step explanation:

the answer will be 3

4 0
3 years ago
Read 2 more answers
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