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ioda
3 years ago
6

What is the slope-intercept form of the function described by this table?

Mathematics
1 answer:
ch4aika [34]3 years ago
3 0
<h3>Answer:</h3>

y = -4x +2

<h3>Step-by-step explanation:</h3>

The slope-intercept form of the equation for a line is conveniently found from the point-slope form. In either case, you need to know the slope. That is ...

... slope = m = (change in y)/(change in x)

For the first two points, the slope is ...

... m = (-6 -(-2))/(2 -1) = -4

Using the point-slope equation for slope m and point (h, k) written as ...

... y = m(x -h) +k

we can substitute m = -4 and (h, k) = (1, -2) to get ...

... y = -4(x -1) -2

... y = -4x +2 . . . . . . . simplify

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8 0
3 years ago
Read 2 more answers
A real estate agent would like to predict the selling price of a single-family house by predicting the price (in thousands of do
forsale [732]

Answer:

$173493.8

Step-by-step explanation:

Data provided in the question:

LSRL for the data is ? = 3.8785x + 18.3538

Here,

x is area in 100 square feet

and

price in thousands of dollar

Thus,

For the given area 4000 square foot

x = 4000 ÷ 100 = 40                  [Area in 100 square feet]

Therefore,

Using the given equation

Price  = 3.8785(40) + 18.3538

or

Price = 173.4938 in thousands of dollar

or

Price = 173.4938 × $1000

Price = $173493.8

8 0
4 years ago
Inverse laplace of L^-1 {5s/s^2 + 3s - 4}​
Salsk061 [2.6K]

Answer:

e^t+e^{-4t}

Step-by-step explanation:

We have to simplify the original function using partial fraction, hence:

\frac{5s}{s^2+3s-4} =\frac{5s}{(s-1)(s+4)}\\\\=\frac{A}{s-1}+\frac{B}{s+4}\\\\Therefore:\\\\\frac{A(s+4)+B(s-1)}{(s-1)(s+4)}=\frac{5s}{(s-1)(s+4)}\\\\Eliminating\ the \ denominator:\\\\A(s+4)+B(s-1)=5s\\\\substitute\ s=1:\\\\A(1+4)+B(1-1)=5(1)\\\\5A=5\\\\A=1\\\\tsubstitute\ s=-4:\\\\A(-4+4)+B(-4-1)=5(-4)\\\\-5B=-20\\\\B=4\\\\Therefore\ substituting\ A\ and\ B\ gives:\\\\\frac{5s}{s^2+3s-4}=\frac{1}{s-1}+ \frac{4}{s+4}\\\\

From\ Laplace\ inverse:\\\\But\ L^{-1}[\frac{1}{s-a} ]=e^{at}\\\\Hence:\\\\L^{-1} [\frac{5s}{s^2+3s-4}]=L^{-1}[\frac{1}{s-1} ]+L^{-1}[\frac{4}{s+4} ]=e^{t}+4e^{-4t}\\\\L^{-1} [\frac{5s}{s^2+3s-4}]=e^{t}+4e^{-4t}

5 0
3 years ago
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