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3 years ago

What is the slope-intercept equation of the line through the points (-7,2) (-2,1)

Mathematics

1 answer:

klasskru [66]3 years ago

6 0

The answer is -1/5. If you use slope intercept formula of (-7,2) and (-2,1) you get -1/5

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Calculate the missing terms in each proportion.

Solnce55 [7]

Answer:

1:11=11:121

5:2:1=35:14:7

1.5:1.25:2.4=6:5:9.6

Step-by-step explanation:

x/11=11/121 so x equals 1

5/2=x/14 so x equals 35

1.5/x=6/5 so x equals 1.25

1.25:2.4=5/x so x equals 9.6

5 0

3 years ago

The following data was collected to explore how the number of square feet in a house, the number of bedrooms, and the age of the

Daniel [21]

Answer:

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Step-by-step explanation:

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3 years ago

What would justify these triangles????

olganol [36]

Answer:

ASA would justify it to be congruent.

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3 years ago

In this problem, you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+4y=12te^(−2t)−(8t+12) with

Zarrin [17]

First check the characteristic solution:

y'' + 4y' + 4y = 0

has characteristic equation

r ² + 4r + 4 = (r + 2)² = 0

with a double root at r = -2, so the characteristic solution is

$y_c = C_1e^{-2t} + C_2te^{-2t}$

For the particular solution corresponding to $12te^{-2t}$ , we might first try the ansatz

$y_p = (At+B)e^{-2t}$

but $e^{-2t}$ and $te^{-2t}$ are already accounted for in the characteristic solution. So we instead use

$y_p = (At^3+Bt^2)e^{-2t}$

which has derivatives

${y_p}' = (-2At^3+(3A-2B)t^2+2Bt)e^{-2t}$

${y_p}'' = (4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t}$

Substituting these into the left side of the ODE gives

$(4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t} + 4(-2At^3+(3A-2B)t^2+2Bt)e^{-2t} + 4(At^3+Bt^2)e^{-2t} \\\\ = (6At+2B)e^{-2t} = 12te^{-2t}$

so that 6A = 12 and 2B = 0, or A = 2 and B = 0.

For the second solution corresponding to $-8t-12$ , we use

$y_p = Ct + D$

with derivative

${y_p}' = C$

${y_p}'' = 0$

Substituting these gives

$4C + 4(Ct+D) = 4Ct + 4C + 4D = -8t-12$

so that 4C = -8 and 4C + 4D = -12, or C = -2 and D = -1.

Then the general solution to the ODE is

$y = C_1e^{-2t} + C_2te^{-2t} + 2t^3e^{-2t} - 2t - 1$

Given the initial conditions y (0) = -2 and y' (0) = 1, we have

$-2 = C_1 - 1 \implies C_1 = -1$

$1 = -2C_1 + C_2 - 2 \implies C_2 = 1$

and so the particular solution satisfying these conditions is

$y = -e^{-2t} + te^{-2t} + 2t^3e^{-2t} - 2t - 1$

$\boxed{y = (2t^3+t-1)e^{-2t} - 2t - 1}$

7 0

3 years ago

When testing the difference between two population means and the population variances are unknown and unequal, the degrees of fr

levacccp [35]

Answer: 34 degrees of freedom should be used to find the p-value of the test .

Step-by-step explanation:

Degrees of Freedom relates to the maximum number of independent values, that have independence to vary in the sample.

Given : When testing the difference between two population means and the population variances are unknown and unequal, the degrees of freedom are calculated as 34.7.

But degree of freedom must be an integer , so we find the greatest integer less than equal to the calculated degree of freedom.

i.e. [df]=[34.7]= 34

Thus , 34 degrees of freedom should be used to find the p-value of the test .

6 0

4 years ago