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Genrish500 [490]

3 years ago

10

Find an equation for the line that passes through the point (-3, 6), parallel to the line through the points (0,-7) and (4,-15).

Write your answer in point-slope form.

1 answer:

Vitek1552 [10]3 years ago

8 0

Broken down into steps:

1. Find the slope of the line segment that connecs the points (0,-7) and (4,-15).

2. Start with the point-slope formula for the equation of a straight line:

y-b = m(x-a), where the given point is (a,b). Borrow the value of m that you calculated in (1), above, and insert it into this point-slope formula.
Finish up by subst. the x- and y-values in (-3,6) into this formula.

Done!

You could, of course, solve this result for y if you wished.

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Write the monomial in standard form. name it's coefficient and identify its degree.

Hunter-Best [27]

Answer:

$Standard\ Form = {3n^2} m^{-2}$

Step-by-step explanation:

Given

$\frac{2}{3m^2n} * 4.5n^3$

Required

Write in Standard Form

To start with; the two monomials have to be multiplied together;

$\frac{2}{3m^2n} * 4.5n^3$

$Standard\ Form = \frac{2 * 4.5n^3}{3m^2n}$

Split the numerator and the denominator

$Standard\ Form = \frac{2 * 4.5 * n^3}{3 * m^2 * n}$

Multiply Like terms

$Standard\ Form = \frac{9 * n^3}{3 * m^2 * n}$

Divide 9 by 3 to give 3

$Standard\ Form = \frac{3 * n^3}{m^2 * n}$

Divide n³ by n to n²

$Standard\ Form = \frac{3 * n^2}{m^2 }$

Split fraction

$Standard\ Form = {3 * n^2} * \frac{1}{m^2 }$

From laws of indices;

$\frac{1}{a^n} = a^{-n}$

$Standard\ Form = {3 * n^2} * \frac{1}{m^2 }$ becomes

$Standard\ Form = {3 * n^2} * m^{-2}$

Multiply all together

$Standard\ Form = {3n^2} m^{-2}$

5 0

2 years ago

What statements must be true of an equation before you can use quadratic formula?

Alexus [3.1K]

Before you use the quadratic formula, you have to make sure the equation itself is a quadratic and that a and b are not 0.

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3 years ago

Obtain the general solution to the equation. (x^2+10) + xy = 4x=0 The general solution is y(x) = ignoring lost solutions, if any

alukav5142 [94]

Answer:

$y(x)=4+\frac{C}{\sqrt{x^2+10}}$

Step-by-step explanation:

We are given that a differential equation

$(x^2+10)y'+xy-4x=0$

We have to find the general solution of given differential equation

$y'+\frac{x}{x^2+10}y-\frac{4x}{x^2+10}=0$

$y'+\frac{x}{x^2+10}y=4\frac{x}{x^2+10}$

Compare with

$y'+P(x) y=Q(x)$

We get

$P(x)=\frac{x}{x^2+10}$

$Q(x)=\frac{4x}{x^2+10}$

I.F= $e^{\int\frac{x}{x^2+10} dx}=e^{\frac{1}{2}ln(x^2+10)}$

$e^{ln\sqrt(x^2+10)}=\sqrt{x^2+10}$

$y\cdot \sqrt{x^2+10}=\int \frac{4x}{x^2+10}\times \sqrt{x^2+10} dx+C$

$y\cdot \sqrt{x^2+10}=\int \frac{4x}{\sqrt{x^2+10}}+C$

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3 years ago

Dale is standing 2x+7 feet from the base of a tower, the tower is 8x-4 feet tall. What is the distance between dale and the top

V125BC [204]

Answer:

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Step-by-step explanation:

Dale is standing (2x + 7) feet height from the base of the tower.

And the tower is given to be (8x - 4)feet tall.

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