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3 years ago

6

Rational number on a number line

1 answer:

solong [7]3 years ago

6 0

Positive rational numbers are always represented on the right side of the zero on the number line. While negative rational numbers are always represented on the left side of zero on the number line.

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Factor completely, the expression: 2x^3-2x^2-12x

muminat

$2x^3-2x^2-12x=2x(x^2-x-6)=(*)\\\\------------------------------\\x^2-x-6\\\\a=1;\ b=-1;\ c=-6\\\\\Delta=(-1)^2-4\cdot1\cdot(-6)=1+24=25\\\\\sqrt\Delta=\sqrt{25}=5\\\\x_1=\frac{1-5}{2\cdot1}=\frac{-4}{2}=-2;\ x_2=\frac{1+5}{2\cdot1}=\frac{6}{2}=3\\\\x^2-x-6=(x+2)(x-3)\\\\------------------------------\\\\(*)=2x(x+2)(x-3)$

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3 years ago

lubasha [3.4K]

Remember slope formula is y=mx+b m is the slope b is the y intercept

Part a-1+5/0+1 = 4/1 that just equals 4 slope of g(X) is 4

part b y = 4x + b -1 = 4(0) + b -1 = b so y = 4x - 1

the function g(x) has a bigger y intercept (or where it crosses y on a graph) y intercept of g(x) is 3 and the y intercept of f(x) is -1

This really helps if you graph it

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4 years ago

Read 2 more answers

PLS HELP I WILL GIVE BRAINllESTFind the surface area of the rectangular prism.

galben [10]

Answer:

A=2(wl+hl+hw)

Step-by-step explanation:

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3 years ago

Manny has 48 feet of wood. He wants to use all of it to create a border around a garden. The equation 2 l plus 2 w equals 48 can

AVprozaik [17]

2I + 2w = 48
9 feet is the answer.
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2 years ago

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find

9966 [12]

Answer:

<em>The particular integral of given differential equation</em>

<em> </em> $y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))$ <em></em>

<em> General solution of given differential equation</em>

<em> </em> $y = y_{c} + y_{p}$ <em></em>

<em> </em> $Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))$ <em></em>

<em></em>

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

Given Differential equation y'' − 5 y' + 4 y = x

Given equation in operator form

D²y - 5 Dy + 4 y = x

⇒ ( D² - 5 D + 4 ) y =x

⇒ f(D) y = Q

where f(D) = D² - 5 D + 4 and Q(x) = x

<em>The auxiliary equation f(m) =0</em>

<em> m²-5 m + 4 =0</em>

m² - 4 m - m + 4 =0

m ( m -4 ) -1 ( m-4) =0

(m - 1) =0 and ( m-4) =0

<em> m = 1 and m =4</em>

<em>The complementary function </em>

<em></em> $Y_{c} = C_{1} e^{x} + C_{2} e^{4x}$ <em></em>

<u><em>Step(ii)</em></u>:-

<u><em>particular integral</em></u>

<em>Particular integral</em>

<em> </em> $y_{p} = \frac{1}{f(D)} Q(x) = \frac{1}{D^{2} - 5 D + 4} X$ <em></em>

<em>taking common '4' </em>

<em> </em> $= \frac{1}{4(1 + (\frac{D^{2} - 5 D}{4} ))} X$ <em></em>

<em> </em>

<em> </em> $=\frac{1}{4} (1 + (\frac{D^{2} -5D}{4})^{-1} )} X$ <em></em>

<em>applying binomial expression</em>

<em> ( 1 + x )⁻¹ = 1 - x + x² - x³ +..... </em>

<em> </em> $=\frac{1}{4} (1 - (\frac{D^{2} -5D}{4}) +((\frac{D^{2} -5D}{4})^{2} -...} )X$ <em></em>

<em>Now simplifying and we will use notation D = </em> $\frac{dy}{dx}$ <em></em>

<em> </em> $=\frac{1}{4} (x - (\frac{D^{2} -5D}{4})x +((\frac{D^{2} -5D}{4})^{2}(x) -...}$ <em></em>

<em>Higher degree terms are neglected</em>

<em> </em> $=\frac{1}{4} (x - (\frac{ -5 D}{4}) x)$ <em></em>

<em>The particular integral of given differential equation</em>

<em> </em> $y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))$ <em></em>

<u><em>Final answer</em></u><em>:-</em>

<em> General solution of given differential equation</em>

<em> </em> $y = y_{c} + y_{p}$ <em></em>

<em> </em> $Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))$ <em></em>

<em></em>

<em></em>

<em> </em>

<em> </em>

4 0

3 years ago