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3 years ago

lie is solving the quadratic equation by completing the square. 3x2 + 18x – 8 = 0 3x2 + 18x = 8 What should Lie do next?

Mathematics

2 answers:

ra1l [238]3 years ago

8 0

Answer: In the next step, he takes the factor 3 out of the variable terms and divide it to the other side.

Step-by-step explanation:

ruslelena [56]3 years ago

3 0

Divide 18 by 2(9) then square that (81). add 81 to both sides

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Which of the following are the correct properties of slope?

olchik [2.2K]

Answer:

I think one of them may be C and B but I'm not sure

4 0

3 years ago

What is the slope of the line that passes through points (0,7) and (-3, 0)? O A Α. 7 ОВ. 7 O C. 13 OD 3

Luden [163]

Answer:

$m=\frac{7}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Point (0, 7)

Point (-3, 0)

Step 2: Find slope m

Simply plug in the 2 coordinates into the slope formula to find slope m

Substitute [SF]: $m=\frac{0-7}{-3-0}$
Subtract: $m=\frac{-7}{-3}$
Simplify: $m=\frac{7}{3}$

8 0

3 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

5 0

3 years ago

A rectangular vegetable patch has a perimeter of 28 meters and an area of 48 square

aliina [53]

Answer:

Length = 8 meters

Width = 6 meters

Step-by-step explanation:

Given the following information:

Perimeter of a rectangular vegetable patch: 2(L + W) = 28 meters

Area = L × W = 48 m²

We can solve for the dimensions by using both equations.

First, let's use the formula for the perimeter of the rectangular vegetable patch, and isolate one of the given variables:

2(L + W) = 28

Divide both sides by 2:

$\frac{2(L + W)}{2} = \frac{28}{2}$

L + W = 14

Subtract W from both sides to isolate L:

L + W - W = 14 - W

L = 14 - W

Next, we'll take the formula for the area, and substitute the value for the L from our previous step:

A = 48 = L × W

48 = W × (14 - W)

Distribute W into the parenthesis:

48 = 14W - W²

Add W² and subtract 14W to both sides:

W² - 14W + 48 = 14W - 14W - W² + W²

W² - 14W + 48 = 0 ⇒ This represents a quadratic equation in standard form. We can use the coefficient and constant values to solve for its roots.

a = 1, b = -14, and c = 48

Substitute these values into the quadratic equation:

$x = \frac{-b +/- \sqrt{b^{2}-4ac}}{2a}$

$x = \frac{14 +/- \sqrt{(-14)^{2}-4(1)(48)}}{2(1)}$

$x = \frac{14 +/- \sqrt{196-192}}{2}$

$x = \frac{14 +/- \sqrt{4}}{2}$

$x = \frac{14 + 2}{2}$ , $x = \frac{14 - 2}{2}$

x = 8, x = 6

Now, we can substitute these values into the formulas for the perimeter and area to find the true dimensions of the rectangular vegetable patch.

Perimeter: 2(L + W) = 2(8 + 6) = 28 meters

Area = L × W = 8 × 6 = 48 m²

Therefore, the dimensions of the rectangular vegetable patch are:

Length = 8 meters

Width = 6 meters

Dimensions: 8 × 6 meters.

3 0

3 years ago

|x-4|=1 answer and reason fast

erik [133]

It’s 3 because 3-4=1 please tell me if this was helpful

5 0

3 years ago